Answer to Question #275320 in Statistics and Probability for ZARA

Question

A magic bag contains 8 raspberries, 6 watermelons, and 7 bananas. 5 pieces of fruits are taken out from the magic bag randomly without replacement. Find the probability that

(a) 3 raspberries and 2 watermelons are taken

(b) the 5 pieces of fruits taken are all bananas

(c) all of the 5 fruits taken are the same kind

(d) the 5 pieces taken contain at least 4 raspberries. 

Answer 1

a)

$P=\frac{C^3_{8}\cdot C^2_{6}}{C^5_{26}}=\frac{8!6!}{3!5!2!4!}\frac{5!21!}{26!}=0.0128$

b)

$P=\frac{ C^5_{7}}{C^5_{26}}=\frac{7!}{5!2!}\frac{5!21!}{26!}=0.0003$

c)

$P=\frac{C^5_8}{C^5_{26}}+\frac{C^5_6}{C^5_{26}}+\frac{C^5_7}{C^5_{26}}+\frac{1}{C^5_{26}}=\frac{8!}{5!3!}\frac{5!21!}{26!}+6\frac{5!21!}{26!}+\frac{7!}{5!2!}\frac{5!21!}{26!}+\frac{5!21!}{26!}=$

$=0.0009+0.00009+0.0003+0.00001=0.0013$

d)

$P=\frac{C^4_{8}\cdot C^1_{22}}{C^5_{26}}+\frac{C^5_{8}}{C^5_{26}}=22\frac{8!}{4!4!}\frac{5!21!}{26!}+\frac{8!}{5!3!}\frac{5!21!}{26!}=0.0234+0.0009=0.0243$