Maths · Volume 2 · Chapter 11

Samacheer Class 12 Maths - Probability Distributions

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Complete Class 12 Mathematics book back solutions for Probability Distributions with exam-ready answers.

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EXERCISE 11.1EXERCISE 11.14 questions
Q.1Suppose X is the number of tails occurred when three fair coins are tossed once simultaneously. Find the values of the random variable X and number of points in its inverse images.v
Solution

Sample space size = 2^3 = 8. Number of outcomes with k tails = C(3,k). Thus values and counts: 0 tails: C(3,0)=1; 1 tail: C(3,1)=3; 2 tails: C(3,2)=3; 3 tails: C(3,3)=1.

Answer:

X ∈ {0,1,2,3}. |X^{-1}(0)|=1, |X^{-1}(1)|=3, |X^{-1}(2)|=3, |X^{-1}(3)|=1.

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Q.2An urn contains 5 mangoes and 4 apples. Three fruits are taken at random. If the number of apples taken is a random variable, then find the values of the random variable and number of points in its inverse images.v
Solution

Total ways C(9,3)=84. Number with k apples = C(4,k)·C(5,3−k). Compute: k=0: C(4,0)C(5,3)=1·10=10; k=1: 4·10=40; k=2: 6·5=30; k=3:4·1=4.

Answer:

X ∈ {0,1,2,3}. Counts: 0→10, 1→40, 2→30, 3→4 (total 84).

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Q.3Two balls are chosen randomly from an urn containing 6 red and 8 black balls. Suppose that we win `15 for each red ball selected and we lose `10 for each black ball selected. If X denotes the winning amount, find the values of X and number of points in its inverse images.v
Solution

Total ordered/unordered? Using unordered draws: total C(14,2)=91. Number with r red = C(6,r)C(8,2−r). r=2 ⇒ C(6,2)=15 ⇒ win 2·15=30. r=1 ⇒6·8=48 ⇒ 15−10=5. r=0 ⇒C(8,2)=28 ⇒ −20.

Answer:

Possible X: −20, 5, 30. Counts: X=30 (2 red):15 outcomes; X=5 (1 red,1 black):48 outcomes; X=−20 (0 red):28 outcomes (total 91).

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Q.4A six sided die is marked '2' on one face, '3' on two of its faces, and '4' on remaining three faces. The die is thrown twice. If X denotes the total score in two throws, find the values of the random variable and number of points in its inverse images.v
Solution

Let multiplicities m(2)=1,m(3)=2,m(4)=3. For ordered throws, count pairs (a,b) by m(a)m(b). Sum=4:2+2 ⇒1·1=1. 5:2+3 or 3+2 ⇒1·2+2·1=4. 6:2+4,3+3,4+2 ⇒1·3+2·2+3·1=10. 7:3+4,4+3 ⇒2·3+3·2=12. 8:4+4 ⇒3·3=9. Total 36.

Answer:

X ∈ {4,5,6,7,8}. Ordered-counts out of 36: 4→1, 5→4, 6→10, 7→12, 8→9.

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EXERCISE 11.2EXERCISE 11.27 questions
Q.1Three fair coins are tossed simultaneously. Find the probability mass function for number of heads occurred.v
Solution

Total 8 equally likely outcomes. Number with k heads = C(3,k) ⇒ probability C(3,k)/8.

Answer:

P(X=k)=C(3,k)/8 for k=0,1,2,3, i.e. 1/8, 3/8, 3/8, 1/8 respectively.

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Q.2A six sided die is marked '1' on one face, '3' on two of its faces, and '5' on remaining three faces. The die is thrown twice. If X denotes the total score in two throws, find (i) the probability mass function (ii) the cumulative distribution function (iii) P(0 ≤ X < 4) (iv) P(X ≥ 6). (Interpretation: faces values are 1,3,5 with multiplicities 1,2,3.)v
Solution

Multiplicities m(1)=1,m(3)=2,m(5)=3. Count ordered pairs as m(a)m(b): sum 2:1·1=1; 4:1·2+2·1=4; 6:1·3+2·2+3·1=10; 8:2·3+3·2=12; 10:3·3=9. Probabilities count/36. CDF obtained by cumulative sums. P(X≥6)=1 - P(X≤4) =1 -5/36=31/36.

Answer:

(i) Possible sums 2,4,6,8,10 with probabilities 1/36, 4/36, 10/36, 12/36, 9/36 respectively. (ii) CDF: F(x)=0 (x<2); =1/36 for 2≤x<4; =5/36 for 4≤x<6; =15/36 for 6≤x<8; =27/36 for 8≤x<10; =1 for x≥10. (iii) P(0≤X<4)=P(X=2)=1/36. (iv) P(X≥6)=(10+12+9)/36=31/36.

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Q.3Find the probability mass function and cumulative distribution function of number of girl child in families with 4 children, assuming equal probabilities for boys and girls.v
Solution

Binomial(n=4,p=1/2): P(X=k)=C(4,k)(1/2)^4=C(4,k)/16. CDF are cumulative sums of these probabilities.

Answer:

X∈{0,1,2,3,4}, P(X=k)=C(4,k)/16: 1/16,4/16,6/16,4/16,1/16. CDF: F(x)=0 (x<0); 1/16 (0≤x<1); 5/16 (1≤x<2); 11/16 (2≤x<3); 15/16 (3≤x<4); 1 (x≥4).

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Q.4Suppose a discrete random variable can only take the values 0, 1, and 2. The probability mass function is defined by f(0)=k, f(1)=1/2, f(2)=1/2−k, for otherwise 0. Find (i) the value (range) of k (ii) cumulative distribution function (iii) P(X ≥ 1).v
Solution

Normalization: k + 1/2 + (1/2 − k)=1, so any k satisfying nonnegativity: k≥0 and 1/2−k≥0 ⇒ 0≤k≤1/2. CDF from cumulative sums. P(X≥1)=P(1)+P(2)=1−P(0)=1−k.

Answer:

(i) 0 ≤ k ≤ 1/2. (ii) F(x)=0 for x<0; =k for 0≤x<1; =k+1/2 for 1≤x<2; =1 for x≥2. (iii) P(X≥1)=1−k.

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Q.5The cumulative distribution function of a discrete random variable is F(x)=0 for x<0; F(x)=0.15 for 0≤x<1; F(x)=0.35 for 1≤x<2; F(x)=0.60 for 2≤x<3; F(x)=0.85 for 3≤x<4; F(x)=1 for x≥4. Find (i) the probability mass function p(0),…,p(4) (ii) P(X<3) (iii) P(X≥2).v
Solution

Probabilities are jumps of F: p(k)=F(k)−F(k−). Compute p(0)=0.15, p(1)=0.35−0.15=0.20, p(2)=0.60−0.35=0.25, p(3)=0.85−0.60=0.25, p(4)=1−0.85=0.15. Then P(X<3)=F(2)=0.60; P(X≥2)=1−F(1)=0.65.

Answer:

(i) p(0)=0.15, p(1)=0.20, p(2)=0.25, p(3)=0.25, p(4)=0.15. (ii) P(X<3)=F(2)=0.60. (iii) P(X≥2)=1−F(1)=1−0.35=0.65.

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Q.6A random variable X has the following probability mass function: x=1,2,3,4,5 with f(1)=k/2, f(2)=2k/3, f(3)=2k, f(4)=3k/2, f(5)=k/3. Find (i) k (ii) P(X ≤ 2.5) (iii) P(X < 3).v
Solution

Sum f = k(1/2 + 2/3 + 2 + 3/2 + 1/3) = k(30/6)=5k ⇒5k=1 ⇒k=1/5. Then P(1)=1/10, P(2)=2/15. So P(X≤2)=1/10+2/15=(3+4)/30=7/30. P(X<3) is same.

Answer:

(i) k=1/5. (ii) P(X≤2.5)=P(1)+P(2)=7/30. (iii) P(X<3)=P(1)+P(2)=7/30.

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Q.7The cumulative distribution function of a discrete random variable is F(x)=0 for x<0; F=0.2 for 0≤x<1; F=0.5 for 1≤x<2; F=0.75 for 2≤x<3; F=1 for x≥3. Find (i) the probability mass function (ii) P(X<3) (iii) P(X≥2).v
Solution

p(k)=jump at k: p(0)=0.2; p(1)=0.5−0.2=0.3; p(2)=0.75−0.5=0.25; p(3)=1−0.75=0.25. Then P(X<3)=0.75 and P(X≥2)=1−F(1)=0.5.

Answer:

(i) p(0)=0.2, p(1)=0.3, p(2)=0.25, p(3)=0.25. (ii) P(X<3)=F(2)=0.75. (iii) P(X≥2)=1−F(1)=1−0.5=0.5.

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EXERCISE 11.3EXERCISE 11.36 questions
Q.1The probability density function of X is given by f(x)=k x e^{-x}, x>0 (0 otherwise). Find the value of k.v
Solution

Normalization: ∫_0^∞ k x e^{-x} dx = k·Γ(2) = k·1! = k = 1 ⇒ k=1.

Answer:

k = 1.

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Q.2The probability density function of X is f(x)=x/2 for 0≤x≤2, 0 otherwise. Find (i) P(0.2 ≤ X ≤ 0.6) (ii) P(1.2 ≤ X ≤ 1.8) (iii) P(0.5 ≤ X ≤ 1.5).v
Solution

P(a≤X≤b)=∫_a^b (x/2) dx = (1/4)(b^2−a^2). (i) (1/4)(0.36−0.04)=0.08. (ii) (1/4)(3.24−1.44)=0.45. (iii) (1/4)(2.25−0.25)=0.50.

Answer:

(i) 0.08. (ii) 0.45. (iii) 0.50.

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Q.3Suppose the amount of milk sold daily at a milk booth is distributed with a minimum of 200 litres and a maximum of 600 litres with probability density function f(x)=k x for 200≤x≤600 (0 otherwise). Find (i) k (ii) the distribution function (iii) P(300≤X≤500).v
Solution

Normalize: ∫_{200}^{600} k x dx = k/2(600^2−200^2)=1 ⇒ k = 2/(360000−40000)=1/160000. Then F(x)=∫_{200}^x k t dt = k/2(x^2−200^2). P(300≤X≤500)=F(500)−F(300)=k/2(500^2−300^2)=k·80000=(1/160000)·80000=1/2.

Answer:

(i) k=1/160000. (ii) For x<200:0. For 200≤x≤600: F(x)=k/2 (x^2−200^2). For x>600:1. (iii) P(300≤X≤500)=0.5.

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Q.4The probability density function of X is given by f(x)=k x^3 e^{-x} for x>0 (0 otherwise). Find (i) k (ii) the distribution function (iii) P(X<3) (iv) P(X≤5) (v) P(X≤4).v
Solution

∫_0^∞ x^3 e^{-x} dx = Γ(4)=3! =6, so k·6=1 ⇒ k=1/6. The pdf is Gamma(α=4,β=1). CDF for integer shape 4: F(x)=1−e^{-x}(1 + x + x^2/2 + x^3/6). Evaluate numerically: (iii) at x=3: 1−13 e^{−3}≈0.3528. (iv) at x=5: 1−(1+5+12.5+20.8333)e^{−5}≈0.7348. (v) at x=4: 1−(1+4+8+10.6667)e^{−4}≈0.5665.

Answer:

(i) k=1/6. (ii) F(x)=1−e^{-x}(1+x+x^2/2+x^3/6) for x>0. (iii) P(X<3)=1−13e^{-3}≈0.3528. (iv) P(X≤5)=1−39.3333 e^{-5}≈0.7348. (v) P(X≤4)=1−23.6667 e^{-4}≈0.5665.

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Q.5If X has pdf f(x)=1+x for −1<x<0 and f(x)=1−x for 0≤x<1 (0 otherwise), find (i) the distribution function F(x) (ii) P(−0.3 ≤ X ≤ 0.6).v
Solution

Integrate pdf. For −1≤x<0: F(x)=∫_{−1}^x (1+t)dt = x + x^2/2 +1/2. For 0≤x<1: F(x)=F(0)+∫_0^x (1−t)dt =1/2 + x − x^2/2. Then P=F(0.6)−F(−0.3)= (0.5+0.6−0.18) − (−0.3+0.045+0.5)=0.92−0.245=0.675.

Answer:

(i) F(x)=0 (x<−1); F(x)=x + x^2/2 + 1/2 for −1≤x<0; F(x)=1/2 + x − x^2/2 for 0≤x<1; F(x)=1 (x≥1). (ii) P(−0.3≤X≤0.6)=0.675.

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Q.6If $X$ is a random variable with distribution function $F(x)=0$ for $x<0$, $F(x)=\dfrac{x^2+x}{2}$ for $0\le x<1$, and $F(x)=1$ for $x\ge1$, find (i) the probability density function $f(x)$ and (ii) $P(0.3\le X\le0.6)$.v
Solution

On $0

Answer:

(i) $f(x)=x+\dfrac12,\ 0

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EXERCISE 11.4EXERCISE 11.47 questions
Q.1For the random variable $X$ with the given probability mass function or probability density function as below, find the mean and variance: (i) $f(x)=\frac{1}{10}$ for $x=2,5$ and $f(x)=\frac{1}{5}$ for $x=0,1,3,4$; (ii) $f(x)=\frac{4-x}{6}$ for $x=1,2,3$; (iii) $f(x)=2(x-1)$ for $1<x<2$ and $0$ otherwise; (iv) $f(x)=\frac12e^{-x/2}$ for $x>0$ and $0$ otherwise.v
Solution

(i) The probabilities are $P(0)=P(1)=P(3)=P(4)=\frac15$ and $P(2)=P(5)=\frac1{10}$. Hence $E(X)=0\cdot\frac15+1\cdot\frac15+2\cdot\frac1{10}+3\cdot\frac15+4\cdot\frac15+5\cdot\frac1{10}=\frac{23}{10}$. Also $E(X^2)=0+\frac15+\frac4{10}+\frac9{5}+\frac{16}{5}+\frac{25}{10}=\frac{81}{10}$, so $\mathrm{Var}(X)=\frac{81}{10}-\left(\frac{23}{10}\right)^2=\frac{281}{100}$.

(ii) Here $P(1)=\frac12$, $P(2)=\frac13$, $P(3)=\frac16$. Thus $E(X)=1\cdot\frac12+2\cdot\frac13+3\cdot\frac16=\frac53$ and $E(X^2)=\frac12+\frac43+\frac{9}{6}=\frac{10}{3}$. Therefore $\mathrm{Var}(X)=\frac{10}{3}-\left(\frac53\right)^2=\frac59$.

(iii) $E(X)=\int_1^2 x\,2(x-1)\,dx=\frac53$ and $E(X^2)=\int_1^2 x^2\,2(x-1)\,dx=\frac{17}{6}$. Hence $\mathrm{Var}(X)=\frac{17}{6}-\left(\frac53\right)^2=\frac1{18}$.

(iv) This is an exponential distribution with rate $\lambda=\frac12$. Therefore $E(X)=\frac1\lambda=2$ and $\mathrm{Var}(X)=\frac1{\lambda^2}=4$.

Answer:

(i) mean $=\frac{23}{10}$, variance $=\frac{281}{100}$. (ii) mean $=\frac53$, variance $=\frac59$. (iii) mean $=\frac53$, variance $=\frac1{18}$. (iv) mean $=2$, variance $=4$.

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Q.2Two balls are drawn in succession without replacement from an urn containing four red balls and three black balls. Let X be the possible outcomes drawing red balls. Find the probability mass function and mean for X.v
Solution

Total C(7,2)=21 unordered draws. P(2 reds)=C(4,2)/21=6/21=2/7. P(1 red)=C(4,1)C(3,1)/21=12/21=4/7. P(0)=C(3,2)/21=3/21=1/7. E(X)=0·1/7+1·4/7+2·2/7=(4+4)/7=8/7.

Answer:

X∈{0,1,2}. P(0)=1/7, P(1)=4/7, P(2)=2/7. Mean μ=E(X)=8/7 ≈1.1429.

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Q.3If $\mu$ and $\sigma^2$ are the mean and variance of the discrete random variable $X$, and $E(X+3)=10$ and $E\big((X+3)^2\big)=116$, find $\mu$ and $\sigma^2$.v
Solution

$E(X+3)=E(X)+3=10\Rightarrow E(X)=7$, so $\mu=7$. Variance is unaffected by a shift: $\mathrm{Var}(X)=\mathrm{Var}(X+3)=E\big((X+3)^2\big)-\big(E(X+3)\big)^2=116-10^2=16$. Hence $\sigma^2=16$.

Answer:

$\mu=7,\ \sigma^2=16$.

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Q.4Four fair coins are tossed once. Find the probability mass function, mean and variance for number of heads occurred.v
Solution

Let X be number of heads. X ~ Binomial(n=4,p=1/2). So P(X=k)=C(4,k)(1/2)^k(1/2)^{4-k}=C(4,k)/16 for k=0,...,4. Mean: E[X]=np=4*(1/2)=2. Variance: Var(X)=np(1-p)=4*(1/2)*(1/2)=1.

Answer:

P(X=k)=C(4,k)/16, k=0,1,2,3,4; mean = 2; variance = 1.

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Q.5A commuter train arrives punctually at a station every half hour. Each morning, a student leaves his house to the train station. Let X denote the amount of time, in minutes, that the student waits for the train from the time he reaches the train station. It is known that the pdf of X is f(x)=1/30 for 0<x<30, 0 elsewhere. Obtain and interpret the expected value of X.v
Solution

X ~ Uniform(0,30) with pdf f(x)=1/30 on (0,30). E[X]= (0+30)/2 =15 minutes. Interpretation: if the arrival time of the student is uniform relative to the train schedule, the expected wait is the midpoint of the interval, 15 min.

Answer:

E[X]=15 minutes. Interpretation: average waiting time is 15 minutes.

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Q.6The time to failure in thousands of hours of an electronic equipment used in a manufactured computer has the density function f(x)=3 e^{-3x} for x>0, 0 elsewhere. Find the expected life of this electronic equipment.v
Solution

This is an exponential distribution with rate λ=3. For Exp(λ), E[X]=1/λ =1/3 (thousands of hours). Converting: 1/3×1000 = 1000/3 ≈ 333.33 hours.

Answer:

E[X]=1/3 thousand hours = 1000/3 hours ≈ 333.33 hours.

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Q.7The probability density function of the random variable $X$ is $f(x)=\dfrac{1}{16}\,x\,e^{-x/4}$ for $x>0$ (and $0$ otherwise). Find the mean and variance of $X$.v
Solution

This is a Gamma density with shape $k=2$ and scale $\theta=4$, since $\dfrac{1}{\Gamma(2)\,4^2}=\dfrac1{16}$. For $\mathrm{Gamma}(k,\theta)$: $E[X]=k\theta=2\cdot4=8$ and $\mathrm{Var}(X)=k\theta^2=2\cdot16=32$. (Directly: $E[X]=\frac1{16}\int_0^{\infty}x^2 e^{-x/4}dx=\frac1{16}\cdot2!\cdot4^3=8$; $E[X^2]=\frac1{16}\cdot3!\cdot4^4=96$, so $\mathrm{Var}=96-64=32$.)

Answer:

Mean $=8$, Variance $=32$.

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EXERCISE 11.5EXERCISE 11.59 questions
Q.1For a binomial distribution $B(n,p)$, compute $P(X=k)$ for (i) $n=6,\ p=\tfrac13,\ k=3$; (ii) $n=10,\ p=\tfrac15,\ k=4$; (iii) $n=9,\ p=\tfrac12,\ k=7$.v
Solution

$P(X=k)=\binom nk p^k(1-p)^{n-k}$. (i) $\binom63\left(\tfrac13\right)^3\left(\tfrac23\right)^3=20\cdot\dfrac{8}{729}=\dfrac{160}{729}$. (ii) $\binom{10}{4}\left(\tfrac15\right)^4\left(\tfrac45\right)^6=210\cdot\dfrac{4096}{9765625}=\dfrac{860160}{9765625}$. (iii) $\binom97\left(\tfrac12\right)^9=\dfrac{36}{512}=\dfrac{9}{128}$.

Answer:

(i) $\dfrac{160}{729}\approx0.220$ (ii) $\dfrac{860160}{9765625}\approx0.088$ (iii) $\dfrac{9}{128}\approx0.070$

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Q.2The probability that Mr. Q hits a target at any trial is $\tfrac14$. He tries $10$ times. Find the probability that he hits the target (i) exactly $4$ times (ii) at least once.v
Solution

$X\sim B\!\left(10,\tfrac14\right)$. (i) $P(X=4)=\binom{10}{4}\left(\tfrac14\right)^4\left(\tfrac34\right)^6=210\cdot\dfrac{729}{1048576}\approx0.146$. (ii) $P(X\ge1)=1-P(X=0)=1-\left(\tfrac34\right)^{10}\approx0.944$.

Answer:

(i) $\approx0.146$ (ii) $1-\left(\tfrac34\right)^{10}\approx0.944$

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Q.3Find the mean and variance of $X$ where (i) a fair coin is tossed $100$ times and $X$ is the number of heads; (ii) a fair die is tossed $240$ times and $X$ is the number of times a $4$ appears.v
Solution

For $B(n,p)$: mean $=np$, variance $=npq$. (i) $n=100,\ p=\tfrac12$: mean $=50$, variance $=100\cdot\tfrac12\cdot\tfrac12=25$. (ii) $n=240,\ p=\tfrac16$: mean $=40$, variance $=240\cdot\tfrac16\cdot\tfrac56=\dfrac{100}{3}\approx33.33$.

Answer:

(i) mean $=50$, variance $=25$. (ii) mean $=40$, variance $=\dfrac{100}{3}\approx33.33$.

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Q.4The probability that a certain kind of component survives an electrical test is $\tfrac34$. Find the probability that exactly $3$ of the $5$ components tested survive.v
Solution

$X\sim B\!\left(5,\tfrac34\right)$. $P(X=3)=\binom53\left(\tfrac34\right)^3\left(\tfrac14\right)^2=10\cdot\dfrac{27}{64}\cdot\dfrac1{16}=\dfrac{135}{512}\approx0.264$.

Answer:

$P(X=3)=\dfrac{135}{512}\approx0.264$

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Q.5The defect rate of an electronic device is $5\%$. A random sample of $10$ is taken. Find the probability that (i) at least one is defective (ii) exactly two are defective.v
Solution

$X\sim B(10,0.05)$. (i) $P(X\ge1)=1-(0.95)^{10}\approx1-0.5987=0.4013$. (ii) $P(X=2)=\binom{10}{2}(0.05)^2(0.95)^8=45\cdot0.0025\cdot0.6634\approx0.0746$.

Answer:

(i) $1-(0.95)^{10}\approx0.401$ (ii) $\approx0.075$

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Q.6If the probability that a fluorescent light has a useful life of at least $600$ hours is $0.9$, find the probability that among $12$ such lights (i) exactly $10$ (ii) at least $11$ (iii) at least $2$ will not have a useful life of at least $600$ hours.v
Solution

Let $X\sim B(12,0.9)$ count the lights lasting $\ge600$h. (i) $P(X=10)=\binom{12}{10}(0.9)^{10}(0.1)^2=66\cdot0.3487\cdot0.01\approx0.230$. (ii) $P(X\ge11)=\binom{12}{11}(0.9)^{11}(0.1)+(0.9)^{12}\approx0.377+0.282=0.659$. (iii) Let $Y\sim B(12,0.1)$ count the failures; $P(Y\ge2)=1-(0.9)^{12}-\binom{12}{1}(0.1)(0.9)^{11}\approx1-0.282-0.377=0.341$.

Answer:

(i) $\approx0.230$ (ii) $\approx0.659$ (iii) $\approx0.341$

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Q.7A binomial distribution has mean $6$ and standard deviation $2$. Find the distribution.v
Solution

$np=6$ and $npq=2^2=4$, so $q=\dfrac{npq}{np}=\dfrac46=\dfrac23$, hence $p=\dfrac13$ and $n=\dfrac{6}{1/3}=18$. Thus $X\sim B\!\left(18,\tfrac13\right)$ with $P(X=x)=\binom{18}{x}\left(\tfrac13\right)^x\left(\tfrac23\right)^{18-x}$.

Answer:

$n=18,\ p=\dfrac13,\ q=\dfrac23$; $X\sim B\!\left(18,\tfrac13\right)$.

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Q.8For a binomial variable $X$ with $n=6$ it is given that $4\,P(X=4)=P(X=2)$. Find the distribution, its mean and variance.v
Solution

$4\binom64 p^4q^2=\binom62 p^2q^4$. Since $\binom64=\binom62=15$, this gives $4p^2=q^2$, so $2p=q=1-p$, hence $p=\dfrac13,\ q=\dfrac23$. Mean $=np=6\cdot\tfrac13=2$; variance $=npq=6\cdot\tfrac13\cdot\tfrac23=\dfrac43$.

Answer:

$p=\dfrac13,\ q=\dfrac23$; mean $=2$, variance $=\dfrac43$.

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Q.9For a binomial distribution with $n=5$, $P(X=1)=0.4096$ and $P(X=2)=0.2048$. Find $p$, the mean and the variance.v
Solution

$\dfrac{P(X=2)}{P(X=1)}=\dfrac{\binom52}{\binom51}\cdot\dfrac pq=2\cdot\dfrac pq=\dfrac{0.2048}{0.4096}=\dfrac12$, so $\dfrac pq=\dfrac14$, i.e. $4p=q=1-p$, giving $p=\dfrac15,\ q=\dfrac45$. Mean $=np=5\cdot\tfrac15=1$; variance $=npq=5\cdot\tfrac15\cdot\tfrac45=\dfrac45$.

Answer:

$p=\dfrac15,\ q=\dfrac45$; mean $=1$, variance $=\dfrac45$.

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Choose the CorrectChoose the Correct20 questions
Q.1 Let $X$ be a random variable with probability density function $f(x)=\dfrac{2}{x^3}$ for $x\ge1$ and $0$ otherwise. Which of the following statements is correct?
Answer: Option 2

$E[X]=\int_1^\infty x\cdot\dfrac{2}{x^3}\,dx=\int_1^\infty\dfrac{2}{x^2}\,dx=2$ (finite). But $E[X^2]=\int_1^\infty x^2\cdot\dfrac{2}{x^3}\,dx=\int_1^\infty\dfrac{2}{x}\,dx$ diverges, so the variance does not exist. Hence the mean exists but the variance does not.

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Q.2 A rod of length $2l$ is broken into two pieces at random. The probability density function of the shorter piece is $f(x)=\dfrac1l$ for $0
Answer: Option 4

$E[X]=\int_0^l\dfrac{x}{l}\,dx=\dfrac l2$ and $E[X^2]=\int_0^l\dfrac{x^2}{l}\,dx=\dfrac{l^2}{3}$. So $\mathrm{Var}(X)=\dfrac{l^2}{3}-\dfrac{l^2}{4}=\dfrac{l^2}{12}$.

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Q.3 In a game a player tosses a fair six-sided die. If the face $6$ comes up the player wins ₹$36$; otherwise he loses ₹$k^2$, where $k\in\{1,2,3,4,5\}$ is the face shown. The expected amount (in ₹) to win at this game is
Answer: Option 2

$E=\dfrac16(36)+\dfrac16\big(-(1^2+2^2+3^2+4^2+5^2)\big)=6-\dfrac{55}{6}=-\dfrac{19}{6}$.

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Q.4 A six-sided die numbered $1,2,3,4,5,6$ and a four-sided die numbered $1,2,3,4$ are rolled and the sum is determined. Let $X$ denote this sum. The number of elements in the inverse image of $7$ is
Answer: Option 4

$X=7$ arises from $(3,4),(4,3),(5,2),(6,1)$ — four ordered outcomes. So the inverse image of $7$ has $4$ elements.

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Q.5 A random variable $X$ has a binomial distribution with $n=25$ and $p=0.8$. Then the standard deviation of $X$ is
Answer: Option 4

$\sigma=\sqrt{npq}=\sqrt{25\times0.8\times0.2}=\sqrt4=2$.

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Q.6 Let $X$ represent the difference between the number of heads and the number of tails obtained when a coin is tossed $n$ times. Then the possible values of $X$ are
Answer: Option 2

If $i$ heads occur then there are $n-i$ tails, so $X=i-(n-i)=2i-n$ for $i=0,1,\dots,n$.

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Q.7 If $f(x)=\dfrac1{12}$ for $a
Answer: Option 4

For a uniform density, $\int_a^b\dfrac1{12}\,dx=\dfrac{b-a}{12}=1$, so $b-a=12$. The pairs $(0,12),(5,17),(7,19)$ all differ by $12$, but $24-16=8\ne12$. So $16$ and $24$ cannot be the values.

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Q.8 Four buses carry $42,36,34$ and $48$ students respectively ($160$ in all). A student is selected at random; let $X$ be the number of students on that student’s bus. A bus driver is selected at random; let $Y$ be the number of students on that bus. Then $E(X)$ and $E(Y)$ respectively are
Answer: Option 3

$E(X)=\sum\dfrac{n_i}{160}\,n_i=\dfrac{42^2+36^2+34^2+48^2}{160}=\dfrac{6520}{160}=40.75$, while $E(Y)=\dfrac{42+36+34+48}{4}=\dfrac{160}{4}=40$.

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Q.9Two coins are flipped independently. The first lands heads with probability $0.6$ and the second with probability $0.5$. Let $X$ be the total number of heads. Find $E(X)$.v
Solution

Write $X=X_1+X_2$, where $X_1,X_2$ are the Bernoulli indicators for the two coins. By linearity of expectation, $E(X)=E(X_1)+E(X_2)=0.6+0.5=1.1$.

Answer:

$E(X)=1.1$

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Q.10 On a multiple-choice exam with $3$ choices for each of the $5$ questions, the probability that a student gets $4$ or more correct just by guessing is
Answer: Option 1

Guessing gives $p=\dfrac13$, $n=5$. $P(X\ge4)=\binom54\left(\dfrac13\right)^4\dfrac23+\left(\dfrac13\right)^5=\dfrac{10}{243}+\dfrac1{243}=\dfrac{11}{243}$.

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Q.11 If $P(X=0)=1-P(X=1)$ and $E(X)=3\,\mathrm{Var}(X)$, then $P(X=0)$ is
Answer: Option 4

Let $P(X=1)=p$, so $P(X=0)=1-p$ and $X$ takes only the values $0,1$. Then $E(X)=p$ and $\mathrm{Var}(X)=p(1-p)$. From $p=3p(1-p)$ we get $1=3(1-p)$, so $p=\dfrac23$ and $P(X=0)=\dfrac13$.

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Q.12If $X$ is a binomial random variable with expected value $6$ and variance $2.4$, find $P(X=5)$.v
Solution

$np=6$ and $npq=2.4$ give $q=\dfrac{2.4}{6}=0.4$, so $p=0.6$ and $n=\dfrac{6}{0.6}=10$. Hence $P(X=5)=\binom{10}{5}(0.6)^5(0.4)^5=252\times(0.6)^5(0.4)^5\approx0.2007$.

Answer:

$P(X=5)=\dbinom{10}{5}(0.6)^5(0.4)^5\approx0.2007$

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Q.13 The random variable $X$ has probability density function $f(x)=ax+b$ for $0
Answer: Option 1

$\int_0^1(ax+b)\,dx=\dfrac a2+b=1$ and $E(X)=\int_0^1 x(ax+b)\,dx=\dfrac a3+\dfrac b2=\dfrac7{12}$. Solving these gives $a=1$ and $b=\dfrac12$.

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Q.14 Suppose $X$ takes one of the values $0,1,2$. If for some constant $k$, $P(X=i)=k\,P(X=i-1)$ for $i=1,2$ and $P(X=0)=\dfrac17$, then the value of $k$ is
Answer: Option 2

$P(0)=\dfrac17,\ P(1)=\dfrac k7,\ P(2)=\dfrac{k^2}7$. Summing to $1$: $\dfrac{1+k+k^2}{7}=1\Rightarrow k^2+k-6=0\Rightarrow(k-2)(k+3)=0$. The positive root gives $k=2$.

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Q.15 Which of the following is a discrete random variable? I. The number of cars crossing a particular signal in a day. II. The number of customers in a queue to buy train tickets at a moment. III. The time taken to complete a telephone call.
Answer: Option 1

Counts (I and II) take isolated whole-number values, so they are discrete. A time duration (III) varies continuously, so it is not discrete. Hence I and II.

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Q.16 If $f(x)=2x$ for $0\le x\le a$ (and $0$ otherwise) is a probability density function of a random variable, then the value of $a$ is
Answer: Option 1

$\int_0^a 2x\,dx=a^2=1$, so $a=1$.

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Q.17 The probability mass function of a random variable is $f(-2)=k,\ f(-1)=2k,\ f(0)=3k,\ f(1)=4k,\ f(2)=5k$. Then $E(X)$ is equal to
Answer: Option 4

$\sum f=15k=1\Rightarrow k=\dfrac1{15}$. $E(X)=(-2)k+(-1)2k+0+1\cdot4k+2\cdot5k=10k=\dfrac{10}{15}=\dfrac23$.

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Q.18 Let $X$ have a Bernoulli distribution with mean $0.4$. Then the variance of $2X-3$ is
Answer: Option 4

$p=0.4$, so $\mathrm{Var}(X)=pq=0.4\times0.6=0.24$. Then $\mathrm{Var}(2X-3)=2^2\,\mathrm{Var}(X)=4\times0.24=0.96$.

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Q.19 If in $6$ trials $X$ is a binomial variable satisfying $9\,P(X=4)=P(X=2)$, then the probability of success is
Answer: Option 2

$9\binom64 p^4q^2=\binom62 p^2q^4$. Since $\binom64=\binom62=15$, this gives $9p^2=q^2$, so $3p=q=1-p$, hence $4p=1$ and $p=0.25$.

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Q.20 A salesperson sells a computer to $1$ in every $20$ customers who enter the showroom. The probability that he sells a computer to exactly two of the next three customers is
Answer: Option 1

$p=\dfrac1{20}$, $n=3$. $P(X=2)=\binom32\left(\dfrac1{20}\right)^2\left(\dfrac{19}{20}\right)=3\cdot\dfrac{19}{20^3}=\dfrac{57}{20^3}$.

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