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Probability Questions and Answers

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121.

The probability that a number selected at random from the first 50 natural numbers is a composite number is -.

Answer: B

Split the first 50 natural numbers into three groups: the number 1, the primes, and the composites.

\[ \text{primes up to 50} = 15 \]

\[ \text{composites} = 50 - 15 - 1 = 34 \]

\[ P = \frac{34}{50} = \frac{17}{25} \]

1 is neither prime nor composite, so it must be excluded separately - forgetting it gives 35 and \( \tfrac{7}{10} \).

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122.

A coin is tossed five times. What is the probability that there is at least one tail?

Answer: A

Five tosses give \( 2^{5} = 32 \) outcomes. Only one of them - all heads - has no tail, so use the complement.

\[ P(\text{no tail}) = \frac{1}{32} \]

\[ P(\text{at least one tail}) = 1 - \frac{1}{32} = \frac{31}{32} \]

"At least one" is almost always easier through the complement than by counting one, two, three, four and five tails separately.

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123.

If a number is chosen at random from the set {1, 2, 3, ...., 100}, then the probability that the chosen number is a perfect cube is -.

Answer: A

The perfect cubes up to 100 are

\[ 1^{3} = 1,\quad 2^{3} = 8,\quad 3^{3} = 27,\quad 4^{3} = 64 \]

since \( 5^{3} = 125 \) is already too large.

\[ P = \frac{4}{100} = \frac{1}{25} \]

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124.

Out of first 20 natural numbers, one number is selected at random. The probability that it is either an even number or a prime number is -.

Answer: D

Among the first 20 natural numbers:

\[ \text{even} = 10, \qquad \text{prime} = 8 \ (2, 3, 5, 7, 11, 13, 17, 19) \]

Only 2 is both, so

\[ 10 + 8 - 1 = 17 \quad\Longrightarrow\quad P = \frac{17}{20} \]

2 is the only even prime, which is exactly why the overlap here is a single number.

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125.

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

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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126.

If four coins are tossed, the probability of getting two heads and two tails is -.

Answer: A

Four coins give \( 2^{4} = 16 \) outcomes. Choose which two of the four show heads:

\[ {}^{4}C_{2} = 6 \]

\[ P = \frac{6}{16} = \frac{3}{8} \]

The six arrangements are HHTT, HTHT, HTTH, THHT, THTH and TTHH.

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127.

What is the probability that the minute and hour hands of a clock will form an acute angle at any given time?

Answer: C

The amount of time in a period of 12 hours when the hands will form an acute angle will be the same as that for obtuse angle. So, you would think that the probability is 12. But if you take into account the small amounts of time when the hands are aligned ( 0∘), hands are at right angles and hands are facing in opposite directions (180∘), then the probability would be slightly less than 0.5.

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128.

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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129.

If P (A) and P (not A) are complementary events and P (A) = 0.15, then P (not A) =?

Answer: C

Complementary events cover every possibility between them, so their probabilities add to 1.

\[ P(\text{not } A) = 1 - P(A) = 1 - 0.15 = 0.85 \]

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130.

What is the probability of not picking a king if you choose randomly from a pack of 52 cards?

Answer: B

Since there are 4 kings in a deck of 52, the probability of drawing a king is 4/52 = 1/13.

Hence, the probability of not picking a king is 1 - 1/13 = 12/13.

(∵ For an event E, P (E) =1-P (not E)).

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