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}=(15!)/(3!12!)=(13\cdot14\cdot15)/(2\cdot3)=455`

a)

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

b)

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

c)

`\displaystyle\\|A|=\binom{10}{2}\cdot5=(10!)/(2!8!)\cdot5=(9\cdot10)/(2)\cdot5=225\\\\P(A)=(225)/(455)=(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-(24)/(91)=(67)/(91)\approx73.6\%`