From a pack of 52 cards, two cards are drawn together at random. What is the probability of both the cards being kings?
Answer: D
There are four kings in the pack.
\[ P = \frac{{}^{4}C_{2}}{{}^{52}C_{2}} = \frac{6}{1326} = \frac{1}{221} \]
Equivalently \( \tfrac{4}{52} \times \tfrac{3}{51} = \tfrac{1}{221} \), the second fraction reflecting that the first card is not replaced.
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A card is drawn from a pack of 52 cards. The probability of getting a queen of club or a king of heart is:
Answer: C
There is exactly one queen of clubs and one king of hearts, and no card is both.
\[ P = \frac{1}{52} + \frac{1}{52} = \frac{2}{52} = \frac{1}{26} \]
The question names specific cards, not whole ranks, so the count is 2 rather than 8.
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If three coins are tossed simultaneously, than the probability of getting at least two heads, is
Answer: C
Three coins give \( 2^{3} = 8 \) equally likely outcomes.
Those with at least two heads are
\[ \text{HHT},\ \text{HTH},\ \text{THH},\ \text{HHH} \]
\[ P = \frac{4}{8} = \frac{1}{2} \]
"At least two" includes all three, so the HHH case must not be left out; counting only exactly two would give \( \tfrac{3}{8} \).
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The collection of one or more outcomes from an experiment is called
Answer: B
An event is any subset of the sample space - one outcome or a collection of them.
Rolling an even number, for example, is the event \( \{2, 4, 6\} \).
Probability is the number assigned to an event, and a random variable is a function from outcomes to numbers - neither is the collection itself.
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A box has 5 black and 3 green shirts. One shirt is picked randomly and put in another box. The second box has 3 black and 5 green shirts. Now a shirt is picked from second box. What is the probability of it being a black shirt?
Answer: B
The colour of the transferred shirt is unknown, so split into the two cases and add them.
A black shirt is moved (probability \( \tfrac{5}{8} \)). The second box then has 4 black out of 9:
\[ \frac{5}{8} \times \frac{4}{9} = \frac{20}{72} \]
A green shirt is moved (probability \( \tfrac{3}{8} \)). The second box still has 3 black out of 9:
\[ \frac{3}{8} \times \frac{3}{9} = \frac{9}{72} \]
\[ P = \frac{20}{72} + \frac{9}{72} = \frac{29}{72} \]
The second box holds 9 shirts after the transfer, not 8 - the moved shirt is added to it.
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A bag contains 6 black, 7 red and 2 white balls. A ball is drawn from the bag at random. Find the probability that the ball drawn is black or white.
Answer: A
Total number of balls = 15
Number of balls that are either black or white = 8
Hence the number of favourable outcomes of ball drawn being black or white is 8.
We know that, Probability of an event E, P (E) =number of favourable outcomes/total number of outcomes
So, the required probability is 8/15.
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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.
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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From a well-shuffled pack of 52 cards, a card is drawn at random, find the probability that it is a spade.
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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What will be the possibility of drawing a jack or a spade from a well shuffled standard deck of 52 playing cards?
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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Two dice are thrown at the same time. Find the probability of getting different values on both.
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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