Question

i) $6$ boys and $6$ girls sit in a row at random. Find the probability that the boys and girls sit alternately. ii) Find the odds in favor and odds against drawing $2$ kings form a pack of $52$ playing cards.

Answer 1

(i) Total cases $=12!$

Favorable cases :-

$1)BGBGBGBGBGBG=6!\times 6!$ (Each B,G)

or $2)GBGBGBGBGBGB=6!\times 6!$

$\Rightarrow$ probability = $\frac{2\times 6!\times 6!}{12!}$

(ii) Total number of ways of picking

2 cards from a deck of 52 = ${}^{52}{C}_2$

Number of ways of picking 2 kings

from 4 kings = ${}^4{C}_2$

Probability = $\frac{{}^4{C}_2}{{}^{52}{C}_2}=\frac{4\times 3}{2}\times \frac{2}{52\times 51}$

$=\frac{1}{13\times 17}=\frac{1}{221}$

Odds in favor$=1:220$

Odds against$=220:1$