NIMCET 2021 — Mathematics PYQ

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Question

NIMCET 2021 — Mathematics PYQ

NIMCET | Mathematics | 2021

The three thrown of three dice, the probability of throwing triplets not more than twice is

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Explanation

Probability of getting triplet when 3 dice are thrown
$\frac{6}{6 ^{3}} = \frac{1}{36}$
Probability of not getting triplet when 3 dice are thrown
$< / s p an >< b r >< s p an s t y l e = " f o n t - s i z e : 12 pt; f o n t - f ami l y : a r ia l, h e l v e t i c a, s an s - ser i f; " >= 1 - (\frac{1}{36}) = \frac{35}{36}$
Here, we use binomial distribution.
Probability of throwing triplet not more that twice = Probability (0 triplet) + Probability (1 triplet) + Probability (2 triplets) = 1 – Probability of getting 3 triplets
$\Rightarrow 1 -^{3} C_{3} (\frac{1}{36})^{3} \Rightarrow 1 - \frac{1}{36}$

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