Question #275572

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.


1
Expert's answer
2021-12-07T10:23:27-0500
8+6+7=218+6+7=21

(215)=21!5!(215)!=21(20)(19)(18)(17)1(2)(3)(4)(5)\dbinom{21}{5}=\dfrac{21!}{5!(21-5)!}=\dfrac{21(20)(19)(18)(17)}{1(2)(3)(4)(5)}

=20349=20349

(a)


P(3R & 2W)=(83)(62)(215)P(3R\ \&\ 2W)=\dfrac{\dbinom{8}{3}\dbinom{6}{2}}{\dbinom{21}{5}}

=56(15)20349=40969=\dfrac{56(15)}{20349}=\dfrac{40}{969}

(b)


P(5B)=(75)(215)=2120349=1969P(5B)=\dfrac{\dbinom{7}{5}}{\dbinom{21}{5}}=\dfrac{21}{20349}=\dfrac{1}{969}

(c)

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

=(85)(215)+(65)(215)+(75)(215)=\dfrac{\dbinom{8}{5}}{\dbinom{21}{5}}+\dfrac{\dbinom{6}{5}}{\dbinom{21}{5}}+\dfrac{\dbinom{7}{5}}{\dbinom{21}{5}}

=56+6+2120349=8320349=\dfrac{56+6+21}{20349}=\dfrac{83}{20349}

(d)


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

=(84)(131)(215)+(85)(215)=\dfrac{\dbinom{8}{4}\dbinom{13}{1}}{\dbinom{21}{5}}+\dfrac{\dbinom{8}{5}}{\dbinom{21}{5}}

=13(70)+5620349=46969=\dfrac{13(70)+56}{20349}=\dfrac{46}{969}


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