# How do they compute probability here?

Dear Student,

Solution)

When a die is thrown, then probability of getting a six = $\frac{1}{6}$

then, probability of not getting a six = $1-\frac{1}{6}=\frac{5}{6}$

If the man gets a six in the first throw, then

probability of getting a six = $\frac{1}{6}$

If he does not get a six in first throw, but gets a six in second throw, then

probability of getting a six in the second throw = $\frac{5}{6}\times \frac{1}{6}=\frac{5}{36}$

If he does not get a six in the first two throws, but gets in the third throw, then

probability of getting a six in the third throw = $\frac{5}{6}\times \frac{5}{6}\times \frac{1}{6}=\frac{25}{216}$

probability that he does not get a six in any of the three throws = $\frac{5}{6}\times \frac{5}{6}\times \frac{5}{6}=\frac{125}{216}$

In the first throw he gets a six, then he will receive Re 1.

If he gets a six in the second throw, then he will receive Re (1 - 1) = 0

If he gets a six in the third throw, then he will receive Rs(-1 - 1 + 1) = Rs (-1),

that means he will lose Re 1 in this case.

Expected value = $\left(\frac{1}{6}\times 1\right)+\left(\frac{5}{6}\times \frac{1}{6}\right)\times \left(0\right)+\left(\frac{5}{6}\times \frac{5}{6}\times \frac{1}{6}\right)\times \left(-1\right)+\left(\frac{5}{6}\times \frac{5}{6}\times \frac{5}{6}\times \frac{1}{6}\right)\times \left(-3\right)$

$=\frac{1}{6}+0-\frac{25}{216}-\frac{375}{216}\phantom{\rule{0ex}{0ex}}=\frac{1}{6}-\frac{25}{216}-\frac{375}{216}\phantom{\rule{0ex}{0ex}}=\frac{36-25-375}{216}\phantom{\rule{0ex}{0ex}}=\frac{-364}{216}$

So, he will loose Rs.$\frac{-364}{216}$

Regards!

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