Probability Questions and Answers

Q: From a well-shuffled pack of 52 cards, a card is drawn at random, find the probability that it is a spade.

A pack has 13 cards in each of the four suits.\[ P = \frac{13}{52} = \frac{1}{4} \]The four suits are equally represented, so each has probability \( \tfrac{1}{4} \) without any counting.

Q: A pot has 2 white, 6 black, 4 grey and 8 green balls. If one ball is picked randomly from the pot, what is the probability of it being black or green?

The pot holds \( 2 + 6 + 4 + 8 = 20 \) balls.Black and green are mutually exclusive, so their probabilities add:\[ P = \frac{6}{20} + \frac{8}{20} = \frac{14}{20} = \frac{7}{10} \]No ball is both black and green, so no overlap has to be subtracted.

Q: Two cards are drawn together from a pack of 52 cards. The probability that one is a spade and one is a heart, is:

Choose one of the 13 spades and one of the 13 hearts.\[ P = \frac{13 \times 13}{{}^{52}C_{2}} = \frac{169}{1326} = \frac{13}{102} \]The two cards are drawn together, so the pair is unordered and the denominator uses a combination; the numerator needs no factor of 2, since choosing one from each suit already counts each pair once.

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61.

There are 5 green, 6 black and 7 white balls in a bag. A ball is drawn at random from the bag. Find the probability that it is not white.

A.5/18
B.2/13
C.7/18
D.11/18

Answer: D

Given,

Number of green balls = 5

Number of black balls = 6

Number of white balls = 7

Total number of outcomes = 5 + 6 + 7 = 18

There are 18 balls out of which 11 are not white.

⇒ Number of favourable outcomes = 11

Probability of an event, P (E) = =Number of favourable outcomes / Total number of outcomes

⇒ P (ball drawn is not white) = 11/18

Therefore, probability that the ball drawn is not white is 11/18.

Alternate Method:

P (ball drawn is white) = 7/18

We know that,

P (ball drawn is white) + P (ball drawn is not white) = 1

Because the sum of the probability of an event and its complementary event is always 1.

⇒ P (ball drawn is not white) = 1 - P (ball drawn is white) = 1−7/18 = 11/18

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62.

From a well-shuffled pack of 52 cards, a card is drawn at random, find the probability that it is a spade.

A.1/2
B.1/4
C.1/51
D.1/52

Answer: B

A pack has 13 cards in each of the four suits.

\[ P = \frac{13}{52} = \frac{1}{4} \]

The four suits are equally represented, so each has probability \( \tfrac{1}{4} \) without any counting.

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63.

What will be the possibility of drawing a jack or a spade from a well shuffled standard deck of 52 playing cards?

A.4/13
B.1/26
C.1/13
D.17/52

Answer: A

There are 4 jacks and 13 spades, but the jack of spades is in both groups and must not be counted twice.

\[ P(J \cup S) = P(J) + P(S) - P(J \cap S) \]

\[ = \frac{4}{52} + \frac{13}{52} - \frac{1}{52} = \frac{16}{52} = \frac{4}{13} \]

Adding without the correction gives \( \tfrac{17}{52} \) - the last option.

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64.

Two dice are thrown at the same time. Find the probability of getting different values on both.

A.5/6
B.½
C.1/6
D.1/36

Answer: A

Let E be the event of getting different values on both the dice. The complementary event is getting the same value on both, for which there are 6 favorable outcomes: (1,1), (2,2), (3,3), (4,4), (5,5) and (6,6). Thus, P (not E) = 1/6

Thus, P (E) = 1 - P (not E) = 1 - (1/6) = 5/6

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65.

The sum of the probability of an event and non event is :

A.2
B.1
C.none of these.

Answer: B

An event and its complement are exhaustive: between them they cover every possible outcome, and they cannot both happen.

\[ P(E) + P(\bar{E}) = 1 \]

This is why \( P(\bar{E}) = 1 - P(E) \), the form used whenever the complement is easier to count than the event itself.

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66.

Suresh keeps all his socks in a single drawer. He has 24 pairs of white socks and 18 pairs of grey socks. Suresh picks 3 socks randomly. Find the possibility of Suresh choosing a matching pair.

A.1/36
B.1/108
C.7/36
D.1

Answer: D

There are only two colours of sock in the drawer, so consider the worst case.

The first two socks might be one white and one grey. But the third must match one of them, since there is no third colour available.

\[ P = 1 \]

This is the pigeonhole principle: three items sorted into two categories force at least two into the same one. The actual numbers of pairs never enter the argument.

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67.

If two dice are thrown together, the probability of getting an even number on one die and an odd number on the other is -.

A.1/4
B.1/2
C.3/4
D.3/5

Answer: B

There are two ways this can happen: even on the first and odd on the second, or the other way round.

\[ P = \frac{3}{6} \times \frac{3}{6} + \frac{3}{6} \times \frac{3}{6} = \frac{1}{4} + \frac{1}{4} = \frac{1}{2} \]

Counting only one arrangement gives \( \tfrac{1}{4} \) - the first option - but the question does not say which die shows which.

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68.

A pot has 2 white, 6 black, 4 grey and 8 green balls. If one ball is picked randomly from the pot, what is the probability of it being black or green?

A.3/4
B.7/10
C.4/3
D.1/10

Answer: B

The pot holds \( 2 + 6 + 4 + 8 = 20 \) balls.

Black and green are mutually exclusive, so their probabilities add:

\[ P = \frac{6}{20} + \frac{8}{20} = \frac{14}{20} = \frac{7}{10} \]

No ball is both black and green, so no overlap has to be subtracted.

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69.

Two cards are drawn together from a pack of 52 cards. The probability that one is a spade and one is a heart, is:

A.3/20
B.29/34
C.47/100
D.13/102

Answer: D

Choose one of the 13 spades and one of the 13 hearts.

\[ P = \frac{13 \times 13}{{}^{52}C_{2}} = \frac{169}{1326} = \frac{13}{102} \]

The two cards are drawn together, so the pair is unordered and the denominator uses a combination; the numerator needs no factor of 2, since choosing one from each suit already counts each pair once.

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70.

A bag contains 3 red and 2 blue marbles. A marble is drawn at random. The probability of drawing a black ball is :

A.3/5
B.2/5
C.0/5
D.1/5

Answer: C

The bag holds only red and blue marbles, so a black one cannot be drawn at all.

\[ P(\text{black}) = \frac{\text{number of black marbles}}{\text{total marbles}} = \frac{0}{5} = 0 \]

An event that cannot happen is an impossible event, and its probability is exactly zero.

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