NCERT Solutions: Class 10 Maths Chapter 14 - Probability

For any event $E$, $P(E)+P(\text{not }E)=1$. An impossible event has probability 0 and a sure event has probability 1. Elementary events exhaust all possible outcomes, so their probabilities add to 1. Therefore every event has probability between 0 and 1 inclusive.

(i) The car starting depends on the condition of the car, so the two outcomes are not equally likely. (ii) A shot depends on the player's skill and situation, so shooting and missing are not equally likely. (iii) If a true-false answer is guessed at random, right and wrong are equally likely. (iv) Birth outcomes are not treated as exactly equally likely in real population data, so they need not be equally likely.

In a random toss of an unbiased coin, $P(\text{head})=P(\text{tail})=\dfrac12$. Since neither team is favoured when one team is assigned head and the other tail, the method is fair.

1. A. $\dfrac{2}{3}$
2. B. $-1.5$
3. C. $15\%$
4. D. $0.7$

The probability of any event must lie between 0 and 1 inclusive. The values $\dfrac23$, $15\% =0.15$, and $0.7$ are all in this interval, but $-1.5$ is negative. Therefore $-1.5$ cannot be a probability.

Complementary probabilities add to 1. Thus $P(\text{not }E)=1-P(E)=1-0.05=0.95$.

There are no orange flavoured candies in the bag, so drawing an orange candy is impossible and has probability 0. Since every candy in the bag is lemon flavoured, drawing a lemon candy is certain and has probability 1.

The events 'same birthday' and 'not the same birthday' are complementary. Therefore $P(\text{same birthday})=1-0.992=0.008$.

There are $3+5=8$ balls in all. Favourable outcomes for red $=3$, so $P(\text{red})=\dfrac38$. Not red means black, with 5 favourable outcomes, so $P(\text{not red})=\dfrac58$.

Total marbles $=5+8+4=17$. Thus $P(\text{red})=\dfrac5{17}$ and $P(\text{white})=\dfrac8{17}$. Not green means red or white, so favourable outcomes $=5+8=13$, giving $P(\text{not green})=\dfrac{13}{17}$.

Total coins $=100+50+20+10=180$. Probability of a 50p coin $=\dfrac{100}{180}=\dfrac59$. Coins that are not ₹5 coins $=180-10=170$, so probability $=\dfrac{170}{180}=\dfrac{17}{18}$.

There are $5+8=13$ fish in all, of which 5 are male. Therefore $P(\text{male fish})=\dfrac5{13}$.

There are 8 equally likely outcomes. For 8, there is 1 favourable outcome, so probability $=\dfrac18$. Odd numbers are $1,3,5,7$, so probability $=\dfrac48=\dfrac12$. Numbers greater than 2 are $3,4,5,6,7,8$, so probability $=\dfrac68=\dfrac34$. All numbers $1$ to $8$ are less than 9, so probability $=1$.

A die has outcomes $1,2,3,4,5,6$. Prime outcomes are $2,3,5$, so probability $=\dfrac36=\dfrac12$. Numbers lying between 2 and 6 are $3,4,5$, so probability $=\dfrac36=\dfrac12$. Odd outcomes are $1,3,5$, so probability $=\dfrac36=\dfrac12$.

A deck has 52 cards. There are 2 red kings, so probability $=\dfrac2{52}=\dfrac1{26}$. There are 12 face cards, so $\dfrac{12}{52}=\dfrac3{13}$. There are 6 red face cards, so $\dfrac6{52}=\dfrac3{26}$. There is one jack of hearts and one queen of diamonds, each giving $\dfrac1{52}$. There are 13 spades, so $\dfrac{13}{52}=\dfrac14$.

Initially there are 5 cards and one queen, so $P(\text{queen})=\dfrac15$. If the queen is drawn and put aside, 4 cards remain. One of them is an ace, so $P(\text{ace})=\dfrac14$. No queen remains, so $P(\text{queen})=0$.

Total pens $=12+132=144$. Good pens $=132$. Therefore $P(\text{good pen})=\dfrac{132}{144}=\dfrac{11}{12}$.

(i) There are 4 defective bulbs out of 20, so $P(\text{defective})=\dfrac4{20}=\dfrac15$. (ii) If the first bulb is not defective and is not replaced, then 19 bulbs remain, of which 15 are not defective. Thus the probability is $\dfrac{15}{19}$.

There are 90 equally likely discs. Two-digit numbers from 10 to 90 give 81 outcomes, so probability $=\dfrac{81}{90}=\dfrac9{10}$. Perfect squares up to 90 are $1,4,9,16,25,36,49,64,81$, giving $\dfrac9{90}=\dfrac1{10}$. Multiples of 5 from 5 to 90 are 18 numbers, so probability $=\dfrac{18}{90}=\dfrac15$.

The die has 6 faces. The letter A appears on 2 faces, so $P(A)=\dfrac26=\dfrac13$. The letter D appears on 1 face, so $P(D)=\dfrac16$.

The rectangle in Fig. 14.6 measures $3$ m by $2$ m, so its area is $6$ m$^2$. The circle has diameter 1 m, so radius $=0.5$ m and area $=\pi(0.5)^2=\dfrac{\pi}{4}$ m$^2$. Probability $=\dfrac{\text{area of circle}}{\text{area of rectangle}}=\dfrac{\pi/4}{6}=\dfrac{\pi}{24}$.

Good pens $=144-20=124$. Therefore $P(\text{she will buy})=\dfrac{124}{144}=\dfrac{31}{36}$. Defective pens $=20$, so $P(\text{she will not buy})=\dfrac{20}{144}=\dfrac5{36}$.

With two dice there are 36 equally likely ordered outcomes. Sum 2 occurs in 1 way, sum 3 in 2 ways, sum 4 in 3 ways, sum 5 in 4 ways, sum 6 in 5 ways, sum 7 in 6 ways, and then the counts decrease symmetrically to 1 way for sum 12. Hence the probabilities are the counts divided by 36. Since different sums have different numbers of favourable outcomes, assigning each sum probability $\dfrac1{11}$ is wrong.

Three coin tosses have $2^3=8$ equally likely outcomes. Hanif wins only with HHH or TTT, so favourable winning outcomes $=2$. Losing outcomes $=8-2=6$. Therefore $P(\text{lose})=\dfrac68=\dfrac34$.

There are $6\times6=36$ equally likely ordered outcomes. For 5 not to come up either time, each throw has 5 choices, so favourable outcomes $=5\times5=25$ and probability $=\dfrac{25}{36}$. The event '5 comes up at least once' is the complement, so probability $=1-\dfrac{25}{36}=\dfrac{11}{36}$.

(i) The equally likely outcomes for two coins are HH, HT, TH and TT. Two heads has probability $\dfrac14$, two tails has probability $\dfrac14$, and one of each has probability $\dfrac24=\dfrac12$. So the three listed categories are not equally likely. (ii) A die has three odd outcomes $1,3,5$ and three even outcomes $2,4,6$. Thus odd and even are equally likely, and $P(\text{odd})=\dfrac36=\dfrac12$.