One card is drawn at random from a pack of 52 cards. What is the probability that the card drawn is a face card (Jack, Queen and King only)?
Answer: B
Each suit has three face cards - jack, queen and king - so there are \( 3 \times 4 = 12 \) in the pack.
\[ P = \frac{12}{52} = \frac{3}{13} \]
The ace is not a face card here, as the question makes explicit; counting it would give \( \tfrac{16}{52} \).
Please log in or register to report this question.
The total events to throw three dice simultaneously is
Answer: D
Each die shows one of 6 faces, and the three dice are independent.
\[ 6^{3} = 216 \]
The rule is \( 6^{n} \) for \( n \) dice; \( 3^{4} = 81 \) and \( 3 \times 6 = 18 \) are the shapes of the two nearest distractors.
Please log in or register to report this question.
The probability of a leap year selected at random contain 53 Sunday is:
Answer: C
A leap year has 366 days, which is
\[ 366 = 52 \times 7 + 2 \]
So it holds 52 complete weeks - and therefore 52 Sundays for certain - plus two extra days. A 53rd Sunday appears only if one of those two is a Sunday.
The two leftover days are one of seven equally likely consecutive pairs:
\[ \{\text{Sun,Mon}\},\ \{\text{Mon,Tue}\},\ \{\text{Tue,Wed}\},\ \{\text{Wed,Thu}\},\ \{\text{Thu,Fri}\},\ \{\text{Fri,Sat}\},\ \{\text{Sat,Sun}\} \]
Two of the seven contain a Sunday, so
\[ P = \frac{2}{7} \]
In a non-leap year there is only one leftover day and the probability drops to \( \tfrac{1}{7} \).
Please log in or register to report this question.
A box has 6 black, 4 red, 2 white and 3 blue shirts. Find the probability of drawing 2 black shirts if they are picked randomly?
Answer: D
\[ P = \frac{{}^{6}C_{2}}{{}^{15}C_{2}} = \frac{15}{105} = \frac{1}{7} \]
Equivalently, drawing one after the other: \( \tfrac{6}{15} \times \tfrac{5}{14} = \tfrac{30}{210} = \tfrac{1}{7} \). The \( \tfrac{6}{15} \) option ignores the second draw entirely.
Please log in or register to report this question.
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.
Please log in or register to report this question.
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.
Please log in or register to report this question.
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)).
Please log in or register to report this question.
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 \( \tfrac{25}{64} \).
Please log in or register to report this question.
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.
Please log in or register to report this question.
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 \( \tfrac{1}{36} \).
Please log in or register to report this question.
