Question

Three yam tubers are chosen at random from 15 tubers of which 5 are spoilt.find the probability that, of the three chosen tuber none is spoilt, all are spoilt, exactly one is spoilt, at least one is spoilt

Answer 1

SOLUTION

GIVEN

Three yam tubers are chosen at random from 15 tubers of which 5 are spoilt.

TO DETERMINE

The probability that, of the three chosen tuber

• none is spoilt

• all are spoilt

• exactly one is spoilt

• at least one is spoilt

EVALUATION

Here it is given that three yam tubers are chosen at random from 15 tubers of which 5 are spoilt.

Total Number of yam tubers = 15

Spoilt yam tubers = 5

Non Spoilt yam tubers = 15 - 5 = 10

(i) The required probability that none is spoilt

$\displaystyle \sf{ = \frac{ {}^{10} C_3}{{}^{15} C_3} }$

$\displaystyle \sf{ = \frac{120}{455} }$

$\displaystyle \sf{ = \frac{24}{91} }$

(ii) The required probability that all are spoilt

$\displaystyle \sf{ = \frac{ {}^{5} C_3}{{}^{15} C_3} }$

$\displaystyle \sf{ = \frac{10}{455} }$

$\displaystyle \sf{ = \frac{2}{91} }$

(iii) The required probability that exactly one is spoilt

$\displaystyle \sf{ = \frac{{}^{5} C_1 \times {}^{10} C_2}{{}^{15} C_3} }$

$\displaystyle \sf{ = \frac{5 \times 45}{455} }$

$\displaystyle \sf{ = \frac{45}{91} }$

(iv) The required probability that at least one is spoilt

= 1 - Probability of none is spoilt

$\displaystyle \sf{ = 1 - \frac{24}{91} }$

$\displaystyle \sf{ = \frac{91 - 24}{91} }$

$\displaystyle \sf{ = \frac{67}{91} }$

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