Samacheer Class 12 Maths Volume 2 - Probability Distributions

Your Progress - Chapter 110% complete

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.

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).

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).

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.

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.

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.

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).

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.

Q.5The cumulative distribution function of a discrete random variable is (reconstructed): 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.

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.

Q.7The cumulative distribution function of a discrete random variable is (reconstructed): 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.

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.

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.

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.

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.

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.

Q.6If X has distribution function F(x)=0 for x<0; F(x)=x^2/2 for 0≤x<1; F(x)=1 for x≥1, find (i) the probability density function f(x) (ii) P(0.3 ≤ X ≤ 0.6). (This reconstructed problem assumes the piecewise form above.)v

Solution

Differentiate F on (0,1): f(x)=x (0<x<1). At x=1 there is an atom of size 1−lim_{x→1−}F(x)=1−1/2=1/2. Then probability requested is the integral of x from 0.3 to 0.6: (x^2/2)|_{0.3}^{0.6}=(0.36−0.09)/2=0.135.

Answer:

(i) For 0<x<1: f(x)=dF/dx = x. There is a point mass P(X=1)=1−F(1−)=1−1/2=1/2. (ii) P(0.3≤X≤0.6)=∫_{0.3}^{0.6} x dx =0.135.

EXERCISE 11.4EXERCISE 11.47 questions

Q.1For the random variable X with the given probability mass function or probability density function below, find the mean and variance. (i) discrete pmf with values x=1,2,3,4 and probabilities 1/10, 2/5, 1/5, 1/4 (ii) discrete pmf x=1,2,3 with probabilities 6/11, 1/11, 2/11 (iii) continuous with f(x)=1/2 (−1<x<1)?? (iv) continuous exponential f(x)=e^{−x}/2 for x>0?? (Notes: OCR was unclear; I give solutions under the assumed pmfs/pdfs.)v

Solution

Computed μ and Var from assumed pmf in (i). For remaining subparts original OCR unclear; please supply exact pmfs/pdfs for (ii)-(iv) if you need them.

Answer:

(i) Under assumed pmf p(1)=1/10,p(2)=2/5,p(3)=1/5,p(4)=1/4: μ=Σxp(x)=1(0.1)+2(0.4)+3(0.2)+4(0.25)=0.1+0.8+0.6+1.0=2.5. Var=E(X^2)−μ^2: E(X^2)=1^2·0.1+4·0.4+9·0.2+16·0.25=0.1+1.6+1.8+4.0=7.5 ⇒ Var=7.5−6.25=1.25. (Other parts need clarified data.)

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.

Q.3If μ and σ^2 are the mean and variance of the discrete random variable X, and E(X)=3.10 and E(X^2)=3.1162 (OCR unclear), find μ and σ^2. (Original OCR ambiguous; solution given under assumption E(X)=3.10 and E(X^2)=3.1162.)v

Solution

Variance formula σ^2 = E(X^2) − [E(X)]^2. Plugging the OCR values gives a negative variance, so the provided moments are inconsistent or mistranscribed. Provide correct E(X) and E(X^2) (or exact fractions) for a valid result.

Answer:

Assuming E(X)=3.10 and E(X^2)=3.1162, μ=3.10, σ^2=E(X^2)−μ^2=3.1162−(3.10)^2=3.1162−9.61=−6.4938 (impossible). Please clarify the given expected values; they appear inconsistent.

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.

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.

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.

Q.7The probability density function of the random variable X is (reconstructed) f(x)= [Gamma form] proportional to x^{3} e^{-x/4}, x>0. Find the mean and variance of X.v

Solution

The given pdf corresponds to a Gamma distribution with shape α=4 and scale θ=4 (i.e. X ~ Gamma(4,4)). For Gamma(α,θ): E[X]=αθ =4·4=16, Var(X)=αθ^2=4·4^2=4·16=64. (Note: constant of normalization assumed so pdf is Gamma(4,4).)

Answer:

Mean = 16, Variance = 64.

EXERCISE 11.5EXERCISE 11.512 questions

Q.1Compute P(X=k) for the binomial distribution B(n,p) for: (i) n=6, p=1/3, k=3; (ii) n=10, p=1/5, k=4; (iii) n=9, p=1/2, k=7.v

Solution

Use binomial pmf P(X=k)=C(n,k)p^k(1-p)^{n-k}. (i) P= C(6,3)(1/3)^3(2/3)^3 = 20*(1/27)*(8/27)=160/729. (ii) P= C(10,4)(1/5)^4(4/5)^6 =210*(1/625)*(4096/15625)=210*(4096)/(9765625) (can leave as binomial form). (iii) P= C(9,7)(1/2)^9 = C(9,2)/512 =36/512 =9/128.

Answer:

(i) C(6,3)(1/3)^3(2/3)^3; (ii) C(10,4)(1/5)^4(4/5)^6; (iii) C(9,7)(1/2)^9 = 36/512 = 9/128.

Q.2The probability that Mr. Q hits a target at any trial is p (text truncated).v

Solution

The statement as given is incomplete. If the hit probability is p, then each trial is Bernoulli(p) and for n independent trials X ~ Binomial(n,p) with pmf P(X=k)=C(n,k)p^k(1-p)^{n-k}. For specific numeric p and n plug into this formula.

Answer:

If hit-probability is p, then for n trials the number of hits X ~ Binomial(n,p).

Q.3Using binomial distribution find the mean and variance of X for the following experiments (i) A fair coin is tossed 100 times, X = number of heads. (ii) A fair die is tossed 240 times, X = number of times that four appeared.v

Solution

(i) For coin p=1/2, n=100: E[X]=np=100·1/2=50, Var(X)=np(1-p)=100·(1/2)·(1/2)=25. (ii) For die p=1/6, n=240: E[X]=np=240·(1/6)=40, Var(X)=np(1-p)=240·(1/6)·(5/6)=240·5/36=1200/36=100/3 ≈33.333.

Answer:

(i) mean = 50, variance = 25. (ii) mean = 40, variance = 100/3 ≈ 33.333.

Q.4Suppose he tries at the target 10 times. Find the probability that he hits the target (i) exactly 4 times (ii) at least one time. (Hit probability p assumed.)v

Solution

Let X~Binomial(n=10,p). (i) P(X=4)=C(10,4)p^4(1-p)^6. (ii) P(at least one hit)=1-P(0)=1-(1-p)^{10}. (If a numeric p is given substitute it.)

Answer:

(i) P(X=4)=C(10,4) p^4(1-p)^6. (ii) P(X≥1)=1-(1-p)^{10}.

Q.4The probability that a certain kind of component will survive an electrical test is 3/4. (text truncated)v

Solution

The text is truncated. If the survival probability is p=3/4 and five components are tested independently, then X~Binomial(5,3/4). For example, probability exactly 3 survive is C(5,3)(3/4)^3(1/4)^2 (see next item).

Answer:

Assuming p=3/4, for n=5 the required probability computations use Binomial(5,3/4).

Q.4Find the probability that exactly 3 of the 5 components tested survive (given survival probability 3/4).v

Solution

X ~ Binomial(n=5,p=3/4). P(X=3)=C(5,3)(3/4)^3(1/4)^2 =10*(27/64)*(1/16)=270/1024=135/512 ≈0.2637.

Answer:

P = C(5,3)(3/4)^3(1/4)^2 = 10*(27/64)*(1/16) = 270/1024 = 135/512 ≈ 0.2637.

Q.5A retailer purchases a certain kind of electronic device from a manufacturer. The manufacturer indicates that the defective rate of the device is 5%. The inspector of the retailer randomly picks 10 items from a shipment. What is the probability that there will be (i) at least one defective item (ii) exactly two defective items?v

Solution

Let X~Binomial(n=10,p=0.05) be number defective. (i) P(X≥1)=1-P(X=0)=1-(0.95)^{10}. (ii) P(X=2)=C(10,2)(0.05)^2(0.95)^8.

Answer:

(i) 1-(0.95)^{10}. (ii) C(10,2)(0.05)^2(0.95)^8.

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

Solution

Let X~Binomial(n=12,p=0.9) = number with life ≥600h. (i) P(X=10)=C(12,10)0.9^{10}0.1^2. (ii) P(X≥11)=P(11)+P(12)=C(12,11)0.9^{11}0.1 + 0.9^{12}. (iii) "At least 2 will not have useful life" means at least 2 failures: Y = number failing ~Binomial(12,0.1). P(Y≥2)=1-[P(Y=0)+P(Y=1)]=1-[0.9^{12} + 12·0.1·0.9^{11}].

Answer:

(i) C(12,10)0.9^{10}0.1^2. (ii) C(12,11)0.9^{11}0.1 + 0.9^{12}. (iii) 1 - [0.9^{12} + 12·0.1·0.9^{11}].

Q.6Find the distribution, mean and standard deviation of X. (context missing in source)v

Solution

The item lacks the description of the experiment or parameter values. To answer, the distribution (e.g. Binomial(n,p) or Poisson(λ) or specified pdf) must be given. For Binomial(n,p): mean=np, SD=√(np(1-p)). Please supply the full problem statement.

Answer:

Cannot determine: the question as given is incomplete. Provide the distribution and moments when full problem statement (n and p or pdf) is available.

Q.7The mean and standard deviation of a binomial variate X are respectively 6 and 2. Find (i) the probability mass function (ii) P(X=3) (iii) P(X≥2).v

Solution

Given E[X]=np=6 and SD=2 ⇒ Var(X)=4 ⇒ np(1-p)=4. From np=6, we get 6(1-p)=4 ⇒ 1-p=2/3 ⇒ p=1/3. Then n = np/p =6/(1/3)=18. (i) PMF: P(X=k)=C(18,k)(1/3)^k(2/3)^{18-k}. (ii) P(X=3)=C(18,3)(1/3)^3(2/3)^{15}. (iii) P(X≥2)=1-P(X=0)-P(X=1) where P(X=0)=(2/3)^{18}, P(X=1)=18(1/3)(2/3)^{17}.

Answer:

(i) n=18, p=1/3 so P(X=k)=C(18,k)(1/3)^k(2/3)^{18-k}. (ii) P(X=3)=C(18,3)(1/3)^3(2/3)^{15}. (iii) P(X≥2)=1-P(X=0)-P(X=1).

Q.8If X ~ Binomial(n,p) such that P(X=4)=4 P(X=2) (text garbled), find n and p. (original statement incomplete)v

Solution

Using P(k)=C(n,k)p^k q^{n-k}, ratio P(4)/P(2) = [C(n,4)/C(n,2)]·p^{2}/q^{2}. Setting this =4 gives an equation in n and p. Without the complete original numerical data or additional condition one cannot produce a unique (n,p). If you supply the full statement I will solve it explicitly.

Answer:

Cannot uniquely solve from the truncated statement. If the condition is P(X=4)=4·P(X=2), one can use ratio formula to get an equation in n and p: [C(n,4)/C(n,2)]·(p^2/q^2)=4. Solve this equation for integer n and 0<p<1.

Q.9In a binomial distribution consisting of 5 independent trials, the probability of 1 and 2 successes are 0.4096 and 0.2048 respectively. Find the mean and variance of the random variable.v

Solution

Let X~Binomial(n=5,p). Given P(1)=5 p q^4 =0.4096 and P(2)=10 p^2 q^3 =0.2048. Ratio P(2)/P(1) = (10 p^2 q^3)/(5 p q^4) = 2p/q = 0.2048/0.4096 = 1/2. So 2p/q = 1/2 ⇒ p/q =1/4 ⇒ p = (1/4)q ⇒ p = (1/4)(1-p) ⇒ 1.25 p = 1/4 ⇒ p = 0.2. Then mean np = 5·0.2 =1. Variance npq =1·0.8 =0.8.

Answer:

Mean = 1, Variance = 0.8.

Choose the CorrectChoose the Correct23 questions

Q.1 Let X be random variable with pdf f(x)=1/x^2 for x ≥ 1 and 0 otherwise. Which statement is correct? (1) both mean and variance exist (2) mean exists but variance does not exist (3) both mean and variance do not exist (4) variance exists but mean does not exist.

Answer: 3

E[X]=∫_{1}^{∞} x·(1/x^2) dx = ∫_{1}^{∞} 1/x dx which diverges → mean does not exist. Hence variance also does not exist. So option (3).

Q.1Unclear/malformed question: "(3) 11 (4) 1" (insufficient information).v

Solution

The provided question text appears to be corrupted and cannot be reconstructed unambiguously. Please provide the original clear statement so I can solve it.

Answer:

Cannot determine — question text is unreadable.

Q.2 A rod of length 2l is broken into two pieces at random. The probability density function of the shorter of the two pieces is f(x)=1/l for 0<x<l, 0 otherwise. The mean and variance of the shorter piece are respectively:

Answer: 3

If X is the shorter piece, its pdf is uniform on (0,l): f(x)=1/l for 0<x<l. Then E[X]=∫_0^l x(1/l)dx = l/2. Var(X)=E[X^2]-(E[X])^2 where E[X^2]=∫_0^l x^2(1/l)dx = l^2/3. Thus Var = l^2/3 - (l/2)^2 = l^2/3 - l^2/4 = l^2/12. So option (3).

Q.2 If for some constant k, P(X=i)=k for i=0,1 and P(X=i)=??? (text corrupted). The value of k is asked (options 1,2,3,4).

Answer: Cannot determine — insufficient data

The question text is incomplete/corrupted and cannot be interpreted unambiguously. Please supply the full statement (all given probabilities) so k can be computed (normalization ∑P(X=i)=1).

Q.3A fair six-sided die is tossed. If the face is 6 the player wins ₹36, otherwise (if face k = 1,2,3,4,5) he loses ₹k^2. The expected amount to win is:v

Solution

E = (1/6)·36 + (1/6)·∑_{k=1}^5 (−k^2) = 6 − (1/6)(1^2+2^2+3^2+4^2+5^2) = 6 − 55/6 = −19/6.

Answer:

Option (2) −19/6

Q.4A 6-sided die (faces 1–6) and a 4-sided die (faces 1–4) are rolled. Let X be the sum. How many ordered outcomes give sum 7 (i.e. |X^{-1}(7)|)?v

Solution

Solve i+j=7 with i∈{1..6}, j∈{1..4}. Solutions: (6,1),(5,2),(4,3),(3,4) — 4 outcomes.

Answer:

Option (4) 4

Q.5A random variable X has binomial distribution with n = 25 and p = 0.8 then standard deviation of X is (1) 6 (2) 4 (3) 3 (4) 2v

Solution

σ = sqrt{np(1−p)} = sqrt{25·0.8·0.2} = sqrt{4} = 2.

Answer:

Option (4) 2

Q.6When a coin is tossed n times, let X = (number of heads) − (number of tails). The possible values of X are:v

If i heads occur then tails = n−i, so X = i − (n−i) = 2i − n for i = 0,1,2,...,n. Thus the possible values are 2i−n for i=0,1,...,n.

Q.7If f(x)=1/12 for a<x<b is a pdf, which pair (a,b) cannot be endpoints?v

Solution

For f(x)=1/12 on (a,b) we need ∫_a^b (1/12) dx = (b−a)/12 =1 ⇒ b−a=12. Differences: 12−0=12, 17−5=12, 19−7=12, 24−16=8. Only (16,24) has length 8, so impossible.

Answer:

Option (4) 16 and 24

Q.8Four buses carry 42, 36, 34 and 48 students (total 160). A student picked uniformly at random: X = size of that student's bus. A driver picked uniformly at random: Y = size of that bus. Find E(X) and E(Y).v

Solution

E(Y) = (42+36+34+48)/4 =160/4=40. E(X)=∑ size·(size/160) = (42^2+36^2+34^2+48^2)/160 =6520/160=163/4=40.75.

Answer:

Option (3) E(X)=163/4 (40.75), E(Y)=40

Q.9Two independent coins: P(H on coin1)=0.6, P(H on coin2)=0.5. Let X = total number of heads. Find E(X).v

Solution

By linearity E(X)=E(X1)+E(X2)=0.6+0.5=1.1.

Answer:

E(X) = 1.1

Q.10On a 5-question multiple-choice test with 3 choices each, the student guesses. Probability of getting 4 or more correct is:v

Solution

P(X≥4)=C(5,4)(1/3)^4(2/3)+C(5,5)(1/3)^5 =5·(1/81)·(2/3)+1/243 =10/243+1/243=11/243.

Answer:

Option (1) 11/243

Q.11Question text missing or garbled; cannot reconstruct.v

Solution

The provided item contains only the fragment "(2) 1" with no full question. Insufficient information to solve.

Answer:

Cannot determine — question incomplete

Q.11Let X be Bernoulli with P(X=1)=p, P(X=0)=1−p. If E(X)=3·Var(X), find P(X=0).v

Solution

E(X)=p, Var(X)=p(1−p). Equation p=3p(1−p). If p=0 ⇒ P(X=0)=1 (trivial). For p≠0 divide by p: 1=3(1−p) ⇒ 1=3−3p ⇒ p=2/3 ⇒ P(X=0)=1−p=1/3.

Answer:

P(X=0)=1/3 (principal nontrivial solution); trivial solution P(X=0)=1 also satisfies

Q.12If X is binomial with mean 6 and variance 2.4, find P(X=5).v

Solution

Let np=6 and np(1−p)=2.4 ⇒ 6(1−p)=2.4 ⇒ p=0.6, n=6/0.6=10. So P(X=5)=C(10,5)0.6^5 0.4^5.

Answer:

P(X=5) = C(10,5)(0.6)^5(0.4)^5

Q.13X has pdf f(x)=ax+b for 0<x<1 (0 otherwise). If E(X)=7/12, find a and b.v

Solution

Normalization: ∫_0^1 (ax+b)dx = a/2 + b =1. Expectation: ∫_0^1 x(ax+b)dx = a/3 + b/2 =7/12. Solve: a/2 + b =1 and a/3 + b/2 =7/12 ⇒ subtract: a=1 ⇒ b=1/2.

Answer:

Option (1) a=1, b=1/2

Q.14Question incomplete/garbled: "Suppose that X takes on one of the values 0, 1, and ..."v

Solution

The statement is truncated and lacks necessary data to solve.

Answer:

Cannot determine — question incomplete

Q.15Which of the following is a discrete random variable? I. Number of cars crossing a signal in a day. II. Number of customers in a queue at a moment. III. Time to complete a telephone call.v

Solution

I and II are counts (discrete). III is a continuous time variable.

Answer:

Option (1) I and II

Q.16If f(x)=a x^2 for 0≤x≤1 (0 otherwise) is a pdf, find a.v

Solution

Require ∫_0^1 a x^2 dx = a/3 =1 ⇒ a=3.

Answer:

Option (3) 3

Q.17PMF: x = −2, −1, 0, 1, 2 with f(x)=k/2, k/3, k, 4k, 5k respectively. Find E(X).v

Solution

Normalization: k(1/2+1/3+1+4+5)=k(65/6)=1 ⇒ k=6/65. E(X)=k[−2·(1/2) + (−1)·(1/3)+0·1 +1·4 +2·5] = (6/65)·(38/3) =76/65.

Answer:

E(X) = 76/65

Q.18X ~ Bernoulli with mean 0.4. Find Var(2X−3).v

Solution

Var(2X−3)=4·Var(X)=4·p(1−p)=4·0.4·0.6=0.96.

Answer:

Option (4) 0.96

Q.19In 6 trials X~Binomial(6,p). If 9·P(X=4)=P(X=2), find p.v

Solution

15·9·p^4(1−p)^2 =15·p^2(1−p)^4 ⇒ 9 p^2 = (1−p)^2 ⇒ 3p=1−p ⇒ p=1/4.

Answer:

Option (2) 0.25

Q.20Sales probability p=1/20. What is probability of exactly 2 sales in next 3 customers?v

Solution

P = C(3,2)(1/20)^2(19/20) = 3·(1/400)·(19/20) =57/8000.

Answer:

Option (1) 57/8000

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Next: Ch.12