# 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 =

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 =

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 =

probability that he does not get a six in any of the three throws =

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 = $\left(\frac{1}{6}×1\right)+\left(\frac{5}{6}×\frac{1}{6}\right)×\left(0\right)+\left(\frac{5}{6}×\frac{5}{6}×\frac{1}{6}\right)×\left(-1\right)+\left(\frac{5}{6}×\frac{5}{6}×\frac{5}{6}×\frac{1}{6}\right)×\left(-3\right)$
$=\frac{1}{6}+0-\frac{25}{216}-\frac{375}{216}\phantom{\rule{0ex}{0ex}}=\frac{1}{6}-\frac{25}{216}-\frac{375}{216}\phantom{\rule{0ex}{0ex}}=\frac{36-25-375}{216}\phantom{\rule{0ex}{0ex}}=\frac{-364}{216}$
So, he will loose Rs.$\frac{-364}{216}$
Regards!

• 1
What are you looking for?