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BITSAT 2012
Exam
BITSAT
Year
2012
Subject
Mathematics
Chapter
Probability

BITSAT 2012 · Mathematics · Probability

Two dice are thrown together 4 times. The probability that both...

Two dice are thrown together 4 times. The probability that both dice will show same numbers twice is -

Options

  1. A13\frac{1}{3}
  2. B2536\frac{25}{36}
  3. C25216\frac{25}{216}
  4. DNone of these

Correct answer

C. 25216\frac{25}{216}

Step-by-step solution

(c) The probability of showing same number by both dice p=636=16p=\frac{6}{36}=\frac{1}{6} In binomial distribution here n=4,r=2,p=n=4, r=2, p=16,q=56 req. probability =(nr)qn−rpr=(42)(56)2(16)2=6(2536)(136)=25216\begin{array}{l} \frac{1}{6}, \mathrm{q}=\frac{5}{6} \\ \text{ req. probability }=\binom{n}{r} q^{n-r} p^{r}=\binom{4}{2}\left(\frac{5}{6}\right)^{2}\left(\frac{1}{6}\right)^{2} \\ =6\left(\frac{25}{36}\right)\left(\frac{1}{36}\right)=\frac{25}{216} \end{array}$$
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