What is the probability of not picking a face card when you draw a card at random from a pack of 52 cards?
Answer: C
There are 12 face cards - jack, queen and king in each of the four suits - so 40 cards are not face cards.
\[ P = \frac{40}{52} = \frac{10}{13} \]
Equivalently \( 1 - \tfrac{12}{52} = 1 - \tfrac{3}{13} = \tfrac{10}{13} \).
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24 cards numbered 1, 2, 3, …., 23, 24 are put in a box and mixed thoroughly. One person draws a card from the box. The probability that the number on the card is divisible by 2 or 3 or both is
Answer: B
The total possible outcomes = 24
Numbers divisible by only 2 are 2, 4, 8, 10, 14, 16, 20, 22 (8 numbers) --- (1)
Numbers divisible only by 3 are 3, 9, 15, 21 (4 numbers) --- (2)
Numbers divisible by both 2 and 3 are 6, 12, 18, 24 (4 numbers) --- (3)
From (1), (2) and (3), we see that the number of favourable outcomes is 16 (i.e., 8 + 4 + 4).
We know that, Probability of an event E, P (E) =number of favourable outcomes/ total number of outcomes
=16/24
=2/3
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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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A card is drawn from a well-shuffled deck of playing cards. Find the probability of drawing a black card which is neither a face card nor an ace?
Answer: C
In each suit, there are 9 cards that are not face cards and ace.
Hence, there will be a total of 18 cards in a deck which are black and are not face cards and ace.
We know that, Probability of an event E, P (E) =number of favourable outcomes/total number of outcomes
Required probability is 18/52 = 9/26.
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Each of letters of the word PILOTS is on separate cards, face down on the table. If you pick a card at random, what is the probability that the letter will be a T or an L?
Answer: B
There are 6 outcomes out of which 2 are favourable (which are, getting T or L).
Probability of an event E, P (E) =number of favourable outcomes/ total number of outcomes
Required probability = 2/6 = 1/3.
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A single die is rolled. The probability of getting 1 or an even number is
Answer: B
The favourable outcomes are 1 together with the even numbers 2, 4 and 6.
\[ \{1, 2, 4, 6\} \rightarrow 4 \text{ outcomes} \]
\[ P = \frac{4}{6} = \frac{2}{3} \]
1 is odd, so it does not overlap with the even numbers and nothing has to be subtracted - counting only three outcomes would leave 1 out.
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A bucket contains 10 brown balls, 8 green balls, and 12 red balls and you pick one randomly without looking. What is the probability that the ball will be brown?
Answer: C
The bucket holds \( 10 + 8 + 12 = 30 \) balls.
\[ P = \frac{10}{30} = \frac{1}{3} \]
The decimal 0.33 is only an approximation of \( \tfrac{1}{3} \), which is the exact value.
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A number is chosen at random among the first 100 natural numbers. Find the probability that the number chosen is prime.
Answer: A
There are 25 primes among the first 100 natural numbers.
\[ P = \frac{25}{100} = \frac{1}{4} \]
The count of 25 is worth remembering: there are 4 primes below 10, 21 more up to 100, and 1 is not among them.
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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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A die is thrown once, the probability of getting a composite number on the die is
Answer: A
Of the faces 1 to 6, the composite numbers are 4 and 6.
\[ P = \frac{2}{6} = \frac{1}{3} \]
2, 3 and 5 are prime, and 1 is neither prime nor composite - including it would give \( \tfrac{1}{2} \).
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