Probability

When a coin is tossed once, the possible outcomes are H and T.

Total number of possible outcomes = 2

For getting a head, the favourable outcome is H.

Number of favourable outcomes = 1

$$\begin{aligned}\text{P(Getting a head)} = \frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}} \\[1em] = \frac{1}{2}\end{aligned}$$

Hence, the probability of getting a head is $\mathbf{\frac{1}{2}}$.

(i) Total number of possible outcomes = 6 (i.e., 1, 2, 3, 4, 5, 6)

For getting an odd number, the favourable outcomes are 1, 3 and 5.

Number of favourable outcomes = 3

$$\begin{aligned}\text{P(Getting an odd number)} = \frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}} \\[1em] = \frac{3}{6} \\[1em] = \frac{1}{2}\end{aligned}$$

Hence, the probability of getting an odd number is $\mathbf{\frac{1}{2}}$.

(ii) Total number of possible outcomes = 6 (i.e., 1, 2, 3, 4, 5, 6)

For getting a composite number, the favourable outcomes are 4 and 6.

Number of favourable outcomes = 2

$$\begin{aligned}\text{P(Getting a composite number)} = \frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}} \\[1em] = \frac{2}{6} \\[1em] = \frac{1}{3}\end{aligned}$$

Hence, the probability of getting a composite number is $\mathbf{\frac{1}{3}}$.

(iii) Total number of possible outcomes = 6 (i.e., 1, 2, 3, 4, 5, 6)

For getting a number less than 5, the favourable outcomes are 1, 2, 3 and 4.

Number of favourable outcomes = 4

$$\begin{aligned}\text{P(Getting a number less than 5)} = \frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}} \\[1em] = \frac{4}{6} \\[1em] = \frac{2}{3}\end{aligned}$$

Hence, the probability of getting a number less than 5 is $\mathbf{\frac{2}{3}}$.

(iv) Total number of possible outcomes = 6 (i.e., 1, 2, 3, 4, 5, 6)

For getting a multiple of 3, the favourable outcomes are 3 and 6.

Number of favourable outcomes = 2

$$\begin{aligned}\text{P(Getting a multiple of 3)} = \frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}} \\[1em] = \frac{2}{6} \\[1em] = \frac{1}{3}\end{aligned}$$

Hence, the probability of getting a multiple of 3 is $\mathbf{\frac{1}{3}}$.

In tossing two coins simultaneously, the possible outcomes are HH, HT, TH and TT.

Total number of possible outcomes = 4

(i) For getting exactly one head, the favourable outcomes are HT and TH.

Number of favourable outcomes = 2

$$\begin{aligned}\text{P(Getting exactly one head)} = \frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}} \\[1em] = \frac{2}{4} \\[1em] = \frac{1}{2}\end{aligned}$$

Hence, the probability of getting exactly one head is $\mathbf{\frac{1}{2}}$.

(ii) For getting at least one tail, the favourable outcomes are HT, TH and TT.

Number of favourable outcomes = 3

$$\begin{aligned}\text{P(Getting at least one tail)} = \frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}} \\[1em] = \frac{3}{4}\end{aligned}$$

Hence, the probability of getting at least one tail is $\mathbf{\frac{3}{4}}$.

(iii) For getting no tail, the favourable outcome is HH.

Number of favourable outcomes = 1

$$\begin{aligned}\text{P(Getting no tail)} = \frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}} \\[1em] = \frac{1}{4}\end{aligned}$$

Hence, the probability of getting no tail is $\mathbf{\frac{1}{4}}$.

(iv) For getting at most one head, the favourable outcomes are HT, TH and TT.

Number of favourable outcomes = 3

$$\begin{aligned}\text{P(Getting at most one head)} = \frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}} \\[1em] = \frac{3}{4}\end{aligned}$$

Hence, the probability of getting at most one head is $\mathbf{\frac{3}{4}}$.

Total number of possible outcomes = Total number of tickets

= 10 + 20

= 30

For getting a prize, the favourable outcomes are the 10 prize tickets.

Number of favourable outcomes = Number of prizes = 10

$$\begin{aligned}\text{P(Getting a prize)} = \frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}} \\[1em] = \frac{10}{30} \\[1em] = \frac{1}{3}\end{aligned}$$

Hence, the probability of getting a prize is $\mathbf{\frac{1}{3}}$.

Total number of possible outcomes = Total number of bulbs = 100

Number of defective bulbs = 8

Number of non-defective bulbs = 100 − 8 = 92

(i) For getting a defective bulb, the favourable outcomes are the 8 defective bulbs.

Number of favourable outcomes = 8

$$\begin{aligned}\text{P(Getting a defective bulb)} = \frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}} \\[1em] = \frac{8}{100} \\[1em] = \frac{2}{25}\end{aligned}$$

Hence, the probability that the bulb drawn is defective is $\mathbf{\frac{2}{25}}$.

(ii) For getting a non-defective bulb, the favourable outcomes are the 92 non-defective bulbs.

Number of favourable outcomes = 92

$$\begin{aligned}\text{P(Getting a non-defective bulb)} = \frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}} \\[1em] = \frac{92}{100} \\[1em] = \frac{23}{25}\end{aligned}$$

Hence, the probability that the bulb drawn is non-defective is $\mathbf{\frac{23}{25}}$.

Total number of possible outcomes = Number of letters in the given word = 11

For getting a vowel, the favourable outcomes are E, A, I, A, I and O.

Number of favourable outcomes = Number of vowels in the given word = 6

$$\begin{aligned}\text{P(Getting a vowel)} = \frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}} \\[1em] = \frac{6}{11}\end{aligned}$$

Hence, the probability that the letter chosen is a vowel is $\mathbf{\frac{6}{11}}$.

Total number of possible outcomes = Total number of cards = 19

(i) For getting a prime number, the favourable outcomes are 2, 3, 5, 7, 11, 13, 17 and 19.

Number of favourable outcomes = 8

$$\begin{aligned}\text{P(Getting a prime number)} = \frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}} \\[1em] = \frac{8}{19}\end{aligned}$$

Hence, the probability of getting a prime number is $\mathbf{\frac{8}{19}}$.

(ii) For getting a multiple of 4, the favourable outcomes are 4, 8, 12 and 16.

Number of favourable outcomes = 4

$$\begin{aligned}\text{P(Getting a multiple of 4)} = \frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}} \\[1em] = \frac{4}{19}\end{aligned}$$

Hence, the probability of getting a multiple of 4 is $\mathbf{\frac{4}{19}}$.

(iii) For getting a composite number less than 10, the favourable outcomes are 4, 6, 8 and 9.

Number of favourable outcomes = 4

$$\begin{aligned}\text{P(Getting a composite number less than 10)}\\[1em] = \frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}} \\[1em] = \frac{4}{19}\end{aligned}$$

Hence, the probability of getting a composite number less than 10 is $\mathbf{\frac{4}{19}}$.

(iv) Numbers from 1 to 19 divisible by 3 are 3, 6, 9, 12, 15 and 18.

Numbers from 1 to 19 divisible by 5 are 5, 10 and 15.

So, the numbers divisible by 3 or by 5 are 3, 5, 6, 9, 10, 12, 15 and 18, which are 8 in number.

∴ Numbers neither divisible by 3 nor by 5 = 19 − 8 = 11 (i.e., 1, 2, 4, 7, 8, 11, 13, 14, 16, 17, 19)

For getting a number neither divisible by 3 nor by 5, the favourable outcomes are 1, 2, 4, 7, 8, 11, 13, 14, 16, 17 and 19.

Number of favourable outcomes = 11

$$\begin{aligned}\text{P(Getting a number neither divisible by 3 nor by 5)} \\[1em] = \frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}} \\[1em] = \frac{11}{19}\end{aligned}$$

Hence, the probability of getting a number neither divisible by 3 nor by 5 is $\mathbf{\frac{11}{19}}$.

Total number of possible outcomes = Total number of cards = 52

(i) For getting a black king, the favourable outcomes are the king of spades and the king of clubs.

Number of favourable outcomes = 2

$$\begin{aligned}\text{P(Getting a black king)} = \frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}} \\[1em] = \frac{2}{52} \\[1em] = \frac{1}{26}\end{aligned}$$

Hence, the probability that the card drawn is a black king is $\mathbf{\frac{1}{26}}$.

(ii) There is one jack in each of the four suits.

For getting a jack, the favourable outcomes are the 4 jacks.

Number of favourable outcomes = 4

$$\begin{aligned}\text{P(Getting a jack)} = \frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}} \\[1em] = \frac{4}{52} \\[1em] = \frac{1}{13}\end{aligned}$$

Hence, the probability that the card drawn is a jack is $\mathbf{\frac{1}{13}}$.

(iii) There are 13 cards of the suit of spades.

For getting a spade, the favourable outcomes are the 13 spades.

Number of favourable outcomes = 13

$$\begin{aligned}\text{P(Getting a spade)} = \frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}} \\[1em] = \frac{13}{52} \\[1em] = \frac{1}{4}\end{aligned}$$

Hence, the probability that the card drawn is a spade is $\mathbf{\frac{1}{4}}$.

(iv) There are 4 kings and 4 queens in a pack of 52 cards.

For getting a king or a queen, the favourable outcomes are the 4 kings and 4 queens.

Number of favourable outcomes = 4 + 4 = 8

$$\begin{aligned}\text{P(Getting a king or a queen)} = \frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}} \\[1em] = \frac{8}{52} \\[1em] = \frac{2}{13}\end{aligned}$$

Hence, the probability that the card drawn is a king or a queen is $\mathbf{\frac{2}{13}}$.

Total number of possible outcomes = 6 (i.e., 1, 2, 3, 4, 5, 6)

The even numbers on a die are 2, 4 and 6, of which the composite numbers are 4 and 6.

For getting an even composite number, the favourable outcomes are 4 and 6.

Number of favourable outcomes = 2

$$\begin{aligned}\text{P(Getting an even composite number)} = \\[1em] \frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}} \\[1em] = \frac{2}{6} \\[1em] = \frac{1}{3}\end{aligned}$$

Hence, option 2 is the correct option.

The wheel is divided into 8 equal sectors marked with the letters a, a, b, b, e, g, i and u.

Total number of possible outcomes = 8

For getting a vowel, the favourable outcomes are a, a, e, i and u.

Number of favourable outcomes = 5

$$\begin{aligned}\text{P(Getting a vowel)} = \frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}} \\[1em] = \frac{5}{8}\end{aligned}$$

Hence, option 4 is the correct option.

Total number of possible outcomes = Total number of students = 35

For choosing a girl, the favourable outcomes are the 15 girls.

Number of favourable outcomes = Number of girls = 15

$$\begin{aligned}\text{P(Choosing a girl)} = \frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}} \\[1em] = \frac{15}{35} \\[1em] = \frac{3}{7}\end{aligned}$$

Hence, option 2 is the correct option.

In tossing two coins simultaneously, the possible outcomes are HH, HT, TH and TT.

Total number of possible outcomes = 4

For getting one head and one tail, the favourable outcomes are HT and TH.

Number of favourable outcomes = 2

$$\begin{aligned}\text{P(Getting one head and one tail)} = \frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}} \\[1em] = \frac{2}{4} \\[1em] = \frac{1}{2}\end{aligned}$$

Hence, option 2 is the correct option.

Total number of possible outcomes = Total number of balls in the bag

= 4 + 6

= 10

For getting a black ball, the favourable outcomes are the 4 black balls.

Number of favourable outcomes = Number of black balls = 4

$$\begin{aligned}\text{P(Getting a black ball)} = \frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}} \\[1em] = \frac{4}{10} \\[1em] = \frac{2}{5}\end{aligned}$$

Hence, option 3 is the correct option.

Total number of possible outcomes = Number of letters in the given word = 9

For getting a vowel, the favourable outcomes are O, U, I and E.

Number of favourable outcomes = Number of vowels in the given word = 4

$$\begin{aligned}\text{P(Getting a vowel)} = \frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}} \\[1em] = \frac{4}{9}\end{aligned}$$

Hence, option 2 is the correct option.

Total number of possible outcomes = Total number of cards = 10

For getting a number less than 7, the favourable outcomes are 1, 2, 3, 4, 5 and 6.

Number of favourable outcomes = 6

$$\begin{aligned}\text{P(Getting a number less than 7)} = \frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}} \\[1em] = \frac{6}{10} \\[1em] = \frac{3}{5}\end{aligned}$$

Hence, option 3 is the correct option.

Total number of possible outcomes = Total number of cards = 52

There is one card numbered 5 in each of the four suits.

For getting a 5, the favourable outcomes are the 4 cards numbered 5.

Number of favourable outcomes = 4

$$\begin{aligned}\text{P(Getting a 5)} = \frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}} \\[1em] = \frac{4}{52} \\[1em] = \frac{1}{13}\end{aligned}$$

Hence, option 1 is the correct option.

Total number of possible outcomes = Total number of balls in the bag

= 6 + 8 + 9 + 4

= 27

1. For getting a blue ball, the favourable outcomes are the 9 blue balls.

Number of favourable outcomes = Number of blue balls = 9

$$\begin{aligned}\text{P(Getting a blue ball)} = \frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}} \\[1em] = \frac{9}{27} \\[1em] = \frac{1}{3}\end{aligned}$$

Hence, option (a) is the correct option.

2. For getting a red or a blue ball, the favourable outcomes are the 6 red and 9 blue balls.

Number of favourable outcomes = 6 + 9 = 15

$$\begin{aligned}\text{P(Getting a red or a blue ball)} = \frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}} \\[1em] = \frac{15}{27} \\[1em] = \frac{5}{9}\end{aligned}$$

Hence, option (c) is the correct option.

3. For getting a non-green ball, the favourable outcomes are the 6 red, 9 blue and 4 yellow balls.

Number of favourable outcomes = 27 − 8 = 19

$$\begin{aligned}\text{P(Getting a non-green ball)} = \frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}} \\[1em] = \frac{19}{27}\end{aligned}$$

Hence, option (c) is the correct option.

4. For getting a non-red ball, the favourable outcomes are the 8 green, 9 blue and 4 yellow balls.

Number of favourable outcomes = 27 − 6 = 21

$$\begin{aligned}\text{P(Getting a non-red ball)} = \frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}} \\[1em] = \frac{21}{27} \\[1em] = \frac{7}{9}\end{aligned}$$

Hence, option (a) is the correct option.

We know that,

The probability of occurrence of an event is always greater than or equal to 0 and less than or equal to 1.

So, reason (R) is true.

In a throw of a die,

Total number of possible outcomes = 6 (i.e., 1, 2, 3, 4, 5, 6)

For getting the number 6, the favourable outcome is 6.

Number of favourable outcomes = 1

P(Getting the number 6) = $\frac{1}{6}$

For getting the number 5, the favourable outcome is 5.

Number of favourable outcomes = 1

P(Getting the number 5) = $\frac{1}{6}$

Since the total number of possible outcomes is 6 in both the cases, the probability of getting the number 5 is $\frac{1}{6}$ and not $\frac{1}{5}$.

So, assertion (A) is false.

Hence, option 4 is the correct option.

We know that,

P(E) = $\frac{\text{number of outcomes favourable to E}}{\text{total number of possible outcomes}}$

So, reason (R) is true.

In a throw of a die,

Total number of possible outcomes = 6 (i.e., 1, 2, 3, 4, 5, 6)

For getting an even number, the favourable outcomes are 2, 4 and 6.

Number of favourable outcomes = 3

P(Getting an even number) = $\frac{3}{6}$ = $\frac{1}{2}$

For getting an odd number, the favourable outcomes are 1, 3 and 5.

Number of favourable outcomes = 3

P(Getting an odd number) = $\frac{3}{6}$ = $\frac{1}{2}$

Since both the probabilities are equal, the probability of getting an even number is the same as that of getting an odd number.

∴ Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).

Hence, option 1 is the correct option.

The formula P(E) = $\frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}}$ can be applied only when all the possible outcomes are equally likely.

The six candidates are not necessarily equally likely to receive votes. Since their probabilities are not given, we cannot assume that the probability of voting for candidate number 3 is $\frac{1}{6}$.

∴ The given statement is false.

Hence, option 2 is the correct option.

The wheel is divided into 6 equal sectors, of which five are marked 3, 4, 5, 7 and 8 and one is missing.

Total number of possible outcomes = 6

Given, P(Getting an odd number) = $\frac{1}{2}$

∴ Number of odd numbers on the wheel = $\frac{1}{2} \times 6$ = 3

The odd numbers among 3, 4, 5, 7 and 8 are 3, 5 and 7, which are already 3 in number.

So, the missing number must be an even number.

Given, P(Getting a prime number) = $\frac{2}{3}$

∴ Number of prime numbers on the wheel = $\frac{2}{3} \times 6$ = 4

The prime numbers among 3, 4, 5, 7 and 8 are 3, 5 and 7, which are 3 in number.

So, the missing number must also be a prime number.

The only number which is both even and prime is 2.

∴ The missing number on the wheel could be 2.

Hence, option 4 is the correct option.

We know that,

The probability of an event always lies between 0 and 1, both inclusive.

5% = $\frac{5}{100}$ = 0.05, which lies between 0 and 1.

0.5 lies between 0 and 1.

$\frac{1}{0.5}$ = 2, which is greater than 1.

$\frac{1}{5}$ = 0.2, which lies between 0 and 1.

Since 2 is greater than 1, it cannot be the probability of an event.

Hence, option 3 is the correct option.

Checking option 1 :

Total number of balls = 4 + 6 = 10

P(Getting a red ball) = $\frac{4}{10}$ = 0.4

So, the bag may contain 4 red balls and 6 green balls.

Checking option 2 :

Total number of balls = 6 + 9 = 15

P(Getting a red ball) = $\frac{6}{15} = \frac{2}{5}$ = 0.4

So, the bag may contain 6 red balls and 9 green balls.

Since both the options give the probability 0.4, the bag may contain either of them.

Hence, option 3 is the correct option.

The box contains 6 crayons of different colours, namely red, yellow, blue, brown, black and green.

On taking 1 crayon from the box, the crayon drawn may be of any one of these 6 colours.

∴ Total number of possible outcomes = 6

Hence, option 2 is the correct option.

Total number of possible outcomes = Total number of tickets sold = 400

For winning the first prize, the favourable outcomes are the 10 tickets bought.

Number of favourable outcomes = Number of tickets bought = 10

$$\begin{aligned}\text{P(Winning the first prize)} = \frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}} \\[1em] = \frac{10}{400}\end{aligned}$$

Hence, option 3 is the correct option.

From the given figure, the circular regions on the board award 2 points, 4 points, 6 points, 8 points and 10 points.

Also, no point is given when an arrow misses the board.

So, in a single shot the possible outcomes are 2 points, 4 points, 6 points, 8 points, 10 points and 0 point (missing the board).

Total number of possible outcomes = 6

For getting 6 points, the favourable outcome is 6 points.

Number of favourable outcomes = 1

$$\begin{aligned}\text{P(Getting 6 points)} = \frac{\text{Number of favourable outcomes}}{\text{Total number of possible outcomes}} \\[1em] = \frac{1}{6}\end{aligned}$$

Hence, option 2 is the correct option.