Probability Questions and Answers

Answer: D

Seat the five girls first, then place the boys in the gaps between them so that no two boys are adjacent.

\[ \text{total arrangements} = 7! = 5040 \]

Five girls create six gaps (including the ends). Choose two gaps and seat the boys in them:

\[ 5! \times {}^{6}C_{2} \times 2! = 120 \times 15 \times 2 = 3600 \]

\[ P = \frac{3600}{5040} = \frac{5}{7} \]

The gap method is the standard way to enforce a "no two together" condition.

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.

Answer: A

Two coins give four outcomes: HH, HT, TH, TT. Three of them contain a tail.

\[ P = \frac{3}{4} \]

Using the complement is quicker: the only outcome with no tail is HH, so

\[ P = 1 - \frac{1}{4} = \frac{3}{4} \]

Answer: D

There are \( 4 + 3 + 5 = 12 \) socks in all.

\[ \text{total ways} = {}^{12}C_{2} = 66 \]

Same colour, counted by colour:

\[ {}^{4}C_{2} + {}^{3}C_{2} + {}^{5}C_{2} = 6 + 3 + 10 = 19 \]

\[ P = \frac{19}{66} \]

The three colours are mutually exclusive, so their counts simply add.

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 \( \tfrac{25}{64} \).

Answer: B

A non-leap year has 365 days:

\[ 365 = 52 \times 7 + 1 \]

So there are 52 Thursdays for certain, plus one leftover day. A 53rd Thursday occurs only if that day is a Thursday, and it is equally likely to be any of the seven days.

\[ P = \frac{1}{7} \]

A leap year has two leftover days and the probability rises to \( \tfrac{2}{7} \).

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.

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 \( \tfrac{1}{36} \).

Answer: C

There are \( 6 + 4 + 2 + 3 = 15 \) shirts, of which 11 are not red. Use the complement.

\[ P(\text{no red}) = \frac{{}^{11}C_{4}}{{}^{15}C_{4}} = \frac{330}{1365} \]

\[ P(\text{at least one red}) = 1 - \frac{330}{1365} = \frac{1035}{1365} = \frac{69}{91} \]

Counting "exactly one, exactly two, exactly three, exactly four" would need four separate terms; the complement needs one.

Answer: A

Choose 2 of the 4 red shirts and 1 of the 3 blue, out of all ways to choose 3 from 15.

\[ P = \frac{{}^{4}C_{2} \times {}^{3}C_{1}}{{}^{15}C_{3}} = \frac{6 \times 3}{455} \]

\[ = \frac{18}{455} \]

The two selections are made together, so the counts multiply; the remaining colours play no part.