Probability — Exercise 25(A)

If A and B are two complementary events then the relation between P(A) and P(B) is :

1. P(A) = P(B)

2. P(A) + P(B) = 0

3. P(A) + P(B) = 1

4. none of these

Answer

We know that,

The probabilities of two complimentary events add up to 1.

∴ P(A) + P(B) = 1

Hence, Option 3 is the correct option.

Out of the vowels of English alphabet, one letter is selected at random. The probability of selecting the letter 'O' is :

1. 1

2. $\dfrac{4}{5}$

3. $\dfrac{1}{26}$

4. $\dfrac{1}{5}$

Answer

Vowels of english alphabet : a, e, i, o, u.

Since, there is one letter 'o' in the vowels.

∴ No. of favourable outcomes = 1

P(selecting the letter 'O') = $\dfrac{\text{No. of favourable outcomes}}{\text{Total no. of possible outcomes}} = \dfrac{1}{5}$.

Hence, Option 4 is the correct option.

When a die is thrown, the probability of getting an even number greater than 4 is :

1. $\dfrac{1}{3}$

2. $\dfrac{2}{3}$

3. $\dfrac{1}{6}$

4. $\dfrac{11}{2}$

Answer

When a dice is thrown, the possible outcomes are 1, 2, 3, 4, 5 and 6.

Even no. greater than 4 is 6.

∴ No. of favourable outcomes = 1

P(getting an even number greater than 4)

= $\dfrac{\text{No. of favourable outcomes}}{\text{Total no. of possible outcomes}} = \dfrac{1}{6}$.

Hence, Option 3 is the correct option.

If a letter is drawn from the letters of English alphabet, then the probability, that it is a letter of the word 'DELHI' is :

1. $\dfrac{1}{5}$

2. $\dfrac{5}{26}$

3. $\dfrac{1}{26}$

4. $\dfrac{21}{26}$

Answer

No. of letters in english alphabet = 26

Letters in the word 'DELHI' are 'D', 'E', 'L', 'H', 'I'.

∴ No. of favourable outcomes = 5

P(drawing a letter of the word 'DELHI')

= $\dfrac{\text{No. of favourable outcomes}}{\text{Total no. of possible outcomes}} = \dfrac{5}{26}$.

Hence, Option 2 is the correct option.

A card is selected at random from a well-shuffled deck of 52 cards. The probability of it being a face card is :

1. $\dfrac{3}{13}$

2. $\dfrac{4}{13}$

3. $\dfrac{6}{13}$

4. $\dfrac{8}{13}$

Answer

No. of face cards in a deck of 52 cards = 12

P(drawing a face card) = $\dfrac{\text{No. of favourable outcomes}}{\text{Total no. of possible outcomes}} = \dfrac{12}{52} = \dfrac{3}{13}$.

Hence, Option 1 is the correct option.

A coin is tossed once. Find the probability of :

(i) getting a tail

(ii) not getting a tail

Answer

In a random experiment of tossing coin once, total number of possible outcomes are 2 which are Head (H) and Tail (T)

(i) Favourable outcome is 'getting a tail'.

∴ Number of favourable outcome = 1.

P(getting a tail) = $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{1}{2}$.

Hence, the probability of getting a tail = $\dfrac{1}{2}$.

(ii) Favourable outcome is 'not getting a tail' or 'we can say getting a head'.

∴ Number of favourable outcome = 1.

P(not getting a tail) = $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{1}{2}$.

Hence, the probability of not getting a tail = $\dfrac{1}{2}$.

A bag contains 3 white, 5 black and 2 red balls, all of the same shape and size. A ball is drawn from the bag without looking into it, find the probability that the ball drawn is :

(i) a black ball.

(ii) a red ball.

(iii) a white ball.

(iv) not a red ball.

(v) not a black ball.

Answer

Total number of balls = 3 + 5 + 2 = 10

So, the total number of possible outcomes = 10

(i) There are 5 black balls.

∴ Number of favourable outcomes = 5

P(getting a black ball) = $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{5}{10} = \dfrac{1}{2}$.

Hence, probability of getting a black ball = $\dfrac{1}{2}$.

(ii) There are 2 red balls.

∴ Number of favourable outcomes = 2.

P(getting a red ball) = $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{2}{10} = \dfrac{1}{5}$.

Hence, probability of getting a red ball = $\dfrac{1}{5}$.

(iii) There are 3 white balls.

∴ Number of favourable outcomes = 3.

P(getting a white ball) = $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{3}{10}$.

Hence, probability of getting a white ball = $\dfrac{3}{10}$.

(iv) There are 2 red balls.

∴ 8 (10 - 2) balls which are not red.

∴ Number of favourable outcomes = 8

Thus, P(not getting a red ball) = $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{8}{10} = \dfrac{4}{5}$.

Hence, the probability of not getting a red ball = $\dfrac{4}{5}$.

(v) There are 3 white + 2 red = 5 balls which are not black

∴ Number of favourable outcomes = 5

Thus, P(not getting a black ball) = $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{5}{10} = \dfrac{1}{2}$.

Hence, the probability of not getting a black ball = $\dfrac{1}{2}$.

In a single throw of a dice, find the probability of getting a number :

(i) greater than 4.

(ii) less than or equal to 4.

(iii) not greater than 4.

Answer

Here, the sample space = {1, 2, 3, 4, 5, 6}

So, No. of possible outcomes = 6

(i) No. greater than 4 = {5, 6}

∴ Number of favourable outcomes = 2

Thus, P(getting a no. greater than 4) = $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{2}{6} = \dfrac{1}{3}$.

Hence, the probability of getting a number greater than 4 = $\dfrac{1}{3}$.

(ii) No. less than or equal to 4 = {1, 2, 3, 4}

∴ Number of favourable outcomes = 4

Thus, P(getting a no. less than or equal to 4) = $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{4}{6} = \dfrac{2}{3}$.

Hence, the probability of getting a number less than or equal to 4 = $\dfrac{2}{3}$.

(iii) Numbers not greater than 4 = {1, 2, 3, 4}

∴ Number of favourable outcomes = 4

Thus, P(getting no. not greater than 4) = $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{4}{6} = \dfrac{2}{3}$.

Hence, the probability of getting a number not greater than 4 = $\dfrac{2}{3}$.

From a well shuffled deck of 52 cards, one card is drawn. Find the probability that the card drawn will :

(i) be a black card.

(ii) not be a red card.

(iii) be a red card.

(iv) be a face card.

(v) be a face card of red colour.

Answer

We know that,

Total number of cards = 52

So, the total number of outcomes = 52

There are 13 cards of each type. The cards of heart and diamond are red in colour. Spade and clubs are black. Hence, there are 26 red cards and 26 black cards.

(i) Number of black cards in a deck = 26 (13 spade + 13 club)

∴ Number of favourable outcomes = 26

P(of drawing a black card) = $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{26}{52} = \dfrac{1}{2}$.

Hence, probability of drawing a black card = $\dfrac{1}{2}$.

(ii) Number of red cards in a deck = 26

∴ Number of non-red (black) cards = 52 - 26 = 26.

∴ Number of favourable outcomes = 26

P(of not drawing a red card) = $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{26}{52} = \dfrac{1}{2}$.

Hence, probability of not drawing a red card = $\dfrac{1}{2}$.

(iii) Number of red cards in a deck = 26.

∴ Number of favourable outcomes = 26

P(of drawing a red card) = $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{26}{52} = \dfrac{1}{2}$.

Hence, probability of drawing a red card = $\dfrac{1}{2}$.

(iv) There are 12 face cards (4 kings, 4 queens and 4 jacks) in a deck.

∴ Number of favourable outcomes = 12

P(of drawing a face card) = $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{12}{52} = \dfrac{3}{13}$.

Hence, probability of drawing a face card = $\dfrac{3}{13}$.

(v) There are 26 red cards in a deck, and 6 of these cards are face cards (2 kings, 2 queens and 2 jacks).

∴ Number of favourable outcomes = 6

P(of drawing a red face card) = $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{6}{52} = \dfrac{3}{26}$.

Hence, probability of drawing a red face card = $\dfrac{3}{26}$.

If A and B are two complementary events then what is the relation between P(A) and P(B)?

Answer

Two complementary events, taken together, include all the outcomes for an experiment and the sum of the probabilities of all outcomes is 1.

∴ P(A) + P(B) = 1

Hence, P(A) + P(B) = 1.

If the probability of happening an event A is 0.46. What will be the probability of not happening of the event A?

Answer

P(A) = 0.46

Let P(A') be the probability of not happening of event A.

Then we know that,

⇒ P(A) + P(A') = 1

⇒ P(A') = 1 - P(A)

⇒ P(A') = 1 - 0.46 = 0.54

Hence, the probability of not happening of event A is 0.54

In a T.T. match between Geeta and Ritu, the probability of the winning of Ritu is 0.73. Find the probability of:

(i) winning of Geeta

(ii) not winning of Ritu.

Answer

(i) Winning of Geeta is a complementary event to winning of Ritu.

∴ P(winning of Ritu) + P(winning of Geeta) = 1

⇒ P(winning of Geeta) = 1 - P(winning of Ritu)

⇒ P(winning of Geeta) = 1 - 0.73

⇒ P(winning of Geeta) = 0.27

Hence, the probability of winning of Geeta = 0.27

(ii) P(not winning of Ritu) = P(winning of Geeta) = 0.27

Hence, the probability of not winning of Ritu = 0.27

In a race between Mahesh and John; the probability that John will loose the race is 0.54. Find the probability of :

(i) winning of Mahesh

(ii) winning of John.

Answer

(i) Since, the race is between Mahesh and John.

We can say that,

P(winning of Mahesh) = P(John loosing the race) = 0.54

Hence, the probability of winning of Mahesh = 0.54

(ii) Winning and loosing of John are complementary events.

∴ P(winning of John) + P(loosing of John) = 1

⇒ P(winning of John) + 0.54 = 1

⇒ P(winning of John) = 1 - 0.54 = 0.46

Hence, the probability of winning of John = 0.46

(i) Write the probability of a sure event.

(ii) Write the probability of an even which is impossible.

(iii) For an event E, write a relation representing the range of values of P(E).

Answer

(i) Probability of sure event = 1.

(ii) Since, impossible event cannot occur. Hence, the probability of an impossible event = 0.

(iii) The probability of an event cannot be less than 0 and greater than 1. Hence, 0 ≤ P(E) ≤ 1.

In a single throw of a dice, find the probability of getting :

(i) 5

(ii) 8

(iii) a number less than 8

(iv) a prime number.

Answer

In a single throw of a dice, the possible outcomes are {1, 2, 3, 4, 5, 6}.

∴ No. of possible outcomes = 6.

(i) No. of favourable outcomes (for getting a 5) = 1

P(getting a 5) = $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{1}{6}$.

Hence, the probability of getting a 5 = $\dfrac{1}{6}$.

(ii) Since, no face as number 8 written on it, there is no outcome favourable to 8.

∴ No. of favourable outcomes = 0.

P(getting a number 8) = $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{0}{6}$ = 0.

Hence, the probability of getting 8 = 0.

(iii) Out of 1, 2, 3, 4, 5, 6, the numbers less than 8 are 1, 2, 3, 4, 5 and 6.

∴ No. of favourable outcomes = 6.

P(getting a number less than 8) = $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{6}{6}$ = 1.

Hence, the probability of getting a number less than 8 = 1.

(iv) Out of 1, 2, 3, 4, 5, 6, the prime numbers are 2, 3, 5.

∴ No. of favourable outcomes = 3.

P(getting a prime number) = $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{3}{6} = \dfrac{1}{2}$.

Hence, the probability of getting a prime number = $\dfrac{1}{2}$.

A dice is thrown once. Find the probability of getting :

(i) an even number

(ii) a number between 3 and 8

(iii) an even number or a multiple of 3.

Answer

In a single throw of dice, the possible outcomes are {1, 2, 3, 4, 5, 6}.

∴ No. of possible outcomes = 6.

(i) Out of 1, 2, 3, 4, 5, 6, the even numbers are 2, 4, 6.

∴ No. of favourable outcomes = 3.

P(getting an even number) = $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{3}{6} = \dfrac{1}{2}$.

Hence, the probability of getting an even number = $\dfrac{1}{2}$.

(ii) Out of 1, 2, 3, 4, 5, 6, the numbers between 3 and 8 are 4, 5, 6.

∴ No. of favourable outcomes = 3.

P(getting a number between 3 and 8) = $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{3}{6} = \dfrac{1}{2}$.

Hence, the probability of getting a number between 3 and 8 = $\dfrac{1}{2}$.

(iii) Out of 1, 2, 3, 4, 5, 6, the numbers that are either an even number or a multiple of 3 are 2, 3, 4, 6.

∴ No. of favourable outcomes = 4.

P(getting either an even number or a multiple of 3) = $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{4}{6} = \dfrac{2}{3}$.

Hence, the probability of getting either an even number or a multiple of 3 = $\dfrac{2}{3}$.

Which of the following cannot be the probability of an event ?

(i) $\dfrac{3}{5}$

(ii) 2.7

(iii) 43%

(iv) -0.6

(v) -3.2

(vi) 0.35

Answer

We know that,

0 ≤ P(Event) ≤ 1.

2.7 cannot be probability of an event as it is greater than 1.

Similarly, -0.6 and -3.2 cannot be as probability cannot be negative.

Hence, (ii), (iv) and (v) cannot be probability of an event.

A bag contains six identical black balls. A child withdraws one ball from the bag without looking into it. What is the probability that he takes out:

(i) a white ball ?

(ii) a black ball ?

Answer

There are six identical black balls.

∴ No. of possible outcomes = 6.

(i) Since, there is no white ball.

∴ No. of favourable outcomes (of getting a white ball) = 0.

P(getting a white ball) = $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{0}{6}$ = 0.

Hence, the probability of getting a white ball = 0.

(ii) There are 6 identical black balls.

∴ No. of favourable outcomes (of getting a black ball) = 6.

P(getting a black ball) = $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{6}{6}$ = 1.

Hence, the probability of getting a black ball = 1.

A single letter is selected at random from the word 'Probability'. Find the probability that it is a vowel.

Answer

The word probability has 11 letters.

∴ No. of possible outcomes = 11.

Vowel letters in 'Probability' are 'O', 'A', 'I', 'I'.

No. of favourable outcomes (of getting a vowel) = 4.

P(getting a vowel) = $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{4}{11}$.

Hence, the probability that letter drawn is a vowel = $\dfrac{4}{11}$.

Ramesh chooses a date at random in January for a party (see the following figure).

Find the probability that he chooses :

(i) a Wednesday

(ii) a Friday

(iii) a Tuesday or a Saturday.

Answer

Since, there are 31 days in January.

∴ No. of possible outcomes = 31.

(i) There is wednesday on 1st, 8th, 15th, 22nd and 29th of January.

∴ No. of favourable outcomes = 5

P(getting a Wednesday) = $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{5}{31}$.

Hence, the probability of getting a Wednesday = $\dfrac{5}{31}$.

(ii) There is friday on 3rd, 10th, 17th, 24th and 31st of January.

∴ No. of favourable outcomes = 5

P(getting a Friday) = $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{5}{31}$.

Hence, the probability of getting a Friday = $\dfrac{5}{31}$.

(iii) There is tuesday on 7th, 14th, 21st and 28th of January and Saturday on 4th, 11th, 18th and 25th of January.

∴ No. of favourable outcomes = 8.

P(getting a Tuesday or Saturday) = $\dfrac{\text{No. of favourable outcomes}}{\text{No. of possible outcomes}} = \dfrac{8}{31}$.

Hence, the probability of getting a Tuesday or Saturday = $\dfrac{8}{31}$.