The total number of events of throwing 10 coins simultaneously is
Answer: A
Each coin can land in 2 ways, and the ten coins are independent.
\[ 2^{10} = 1024 \]
The rule is \( 2^{n} \) for \( n \) coins - not \( 2n \), which would give only 20.
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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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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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In a drawer there are 5 black socks and 3 green socks. Two socks are picked randomly one after the other without replacement. What is the possibility that both the socks are black?
Answer: A
Without replacement, the second draw sees one fewer sock of each relevant kind.
\[ P = \frac{5}{8} \times \frac{4}{7} = \frac{20}{56} = \frac{5}{14} \]
Using \( \tfrac{5}{8} \) twice would assume replacement and give \( \tfrac{25}{64} \).
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The special rule of multiplication of probability, the events must be
Answer: A
The special rule of multiplication states that
\[ P(A \cap B) = P(A) \times P(B) \]
and this holds only when the events are independent - the occurrence of one leaving the probability of the other unchanged.
Mutual exclusivity belongs to the special rule of addition, \( P(A \cup B) = P(A) + P(B) \). The two conditions are quite different: mutually exclusive events with non-zero probabilities are in fact never independent, since one occurring makes the other impossible.
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On rolling a dice 2 times, the sum of 2 numbers that appear on the uppermost face is 8. What is the probability that the first throw of dice yields 4?
Answer: D
The sum is given to be 8, so the sample space shrinks to the outcomes that produce it.
\[ (2,6),\ (3,5),\ (4,4),\ (5,3),\ (6,2) \]
Five outcomes, of which exactly one starts with a 4.
\[ P = \frac{1}{5} \]
Note that \( (4,4) \) is a single outcome, not two. Without the given sum, the probability of "first throw 4 and total 8" would be \( \tfrac{1}{36} \).
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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 bag contains 4 white, 5 red and 6 blue balls. Three balls are drawn at random from the bag. The probability that all of them are red, is:
Answer: C
The bag holds \( 4 + 5 + 6 = 15 \) balls, 5 of them red.
\[ P = \frac{{}^{5}C_{3}}{{}^{15}C_{3}} = \frac{10}{455} = \frac{2}{91} \]
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The sum of the probabilities of all elementary events of an experiment is p, then
Answer: C
The elementary events of an experiment are mutually exclusive and between them cover every possible outcome, so their probabilities must account for the whole.
\[ p = 1 \]
This is the normalisation axiom of probability. Any set of assignments summing to less or more than 1 is not a valid probability distribution.
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What is probability of drawing two clubs from a well shuffled pack of 52 cards?
Answer: B
There are 13 clubs in the pack. Draw two without replacement:
\[ P = \frac{13}{52} \times \frac{12}{51} = \frac{1}{4} \times \frac{12}{51} \]
\[ = \frac{12}{204} = \frac{1}{17} \]
Equivalently \( \dfrac{{}^{13}C_{2}}{{}^{52}C_{2}} = \dfrac{78}{1326} = \dfrac{1}{17} \). The second denominator is 51, not 52, because the first card is not replaced.
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