Question

C) Three yam tubers are chosen at random from 15 tubers of which 5 are spoilt. Find the probability that, of the three chosen tubers: a) none is spoilt b) all are spoilt c) exactly one is spoilt d) at least one is spoilt.
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Answer 1

$\displaystyle\\|\Omega|=\binom{15}{3}=\dfrac{15!}{3!12!}=\dfrac{13\cdot14\cdot15}{2\cdot3}=455$

a)

$\displaystyle\\|A|=\binom{10}{3}=\dfrac{10!}{3!7!}=\dfrac{8\cdot9\cdot10}{2\cdot3}=120\\\\P(A)=\dfrac{120}{455}=\dfrac{24}{91}\approx26.4\%$

b)

$\displaystyle\\|A|=\binom{5}{3}=\dfrac{5!}{3!2!}=\dfrac{4\cdot5}{2}=10\\\\P(A)=\dfrac{10}{455}=\dfrac{2}{91}\approx2.2\%$

c)

$\displaystyle\\|A|=\binom{10}{2}\cdot5=\dfrac{10!}{2!8!}\cdot5=\dfrac{9\cdot10}{2}\cdot5=225\\\\P(A)=\dfrac{225}{455}=\dfrac{45}{91}\approx49.5\%$

d)

$A$ - at least one is spoilt

$A'$ - none is spoilt

$P(A)=1-P(A')$

We calculated $P(A')$ in a).

Therefore

$P(A)=1-\dfrac{24}{91}=\dfrac{67}{91}\approx73.6\%$