Answer to Question #275572 in Statistics and Probability for THEASAMOAH

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

$$8+6+7=21$$

$$\dbinom{21}{5}=\dfrac{21!}{5!(21-5)!}=\dfrac{21(20)(19)(18)(17)}{1(2)(3)(4)(5)}$$

$$=20349$$

(a)

$$P(3R\ \&\ 2W)=\dfrac{\dbinom{8}{3}\dbinom{6}{2}}{\dbinom{21}{5}}$$

$$=\dfrac{56(15)}{20349}=\dfrac{40}{969}$$

(b)

$$P(5B)=\dfrac{\dbinom{7}{5}}{\dbinom{21}{5}}=\dfrac{21}{20349}=\dfrac{1}{969}$$

(c)

$$P(5S)+P(5W)+P(5B)$$

$$=\dfrac{\dbinom{8}{5}}{\dbinom{21}{5}}+\dfrac{\dbinom{6}{5}}{\dbinom{21}{5}}+\dfrac{\dbinom{7}{5}}{\dbinom{21}{5}}$$

$$=\dfrac{56+6+21}{20349}=\dfrac{83}{20349}$$

(d)

$$P(at\ least\ 4R)=P(4R)+P(5R)$$

$$=\dfrac{\dbinom{8}{4}\dbinom{13}{1}}{\dbinom{21}{5}}+\dfrac{\dbinom{8}{5}}{\dbinom{21}{5}}$$

$$=\dfrac{13(70)+56}{20349}=\dfrac{46}{969}$$