Find out the probability

Question

'X' is playing a dice game in which a dice is rolled 5 times.If a number turns up exactly 3 times then the game is won.'X' has thrown the dice 2 times and got number 4 both times.What is the probability that 'X' will win the game?

a) 1/216

b)  75/216

c)   80/216

d)  90/216

Comments

For the last case where last 3 throws are same but not 4, can we write it like this-

$\frac{5}{6}* \frac{1}{6}*\frac{1}{6}$

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Seems correct, but not getting close to any option.

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@UK I think this time I get it correct.

"exactly 3 times" is the key

Answer 1

A dice is rolled 5 times, in first two rolls we get 4, 

If A number turns exactly 3 times then game is won. 

Probability that $X$ win the game,  is possible in two cases

First case:

if we get exactly one 4 in next 3 rolls, 

$= \left(\frac{1}{6}*\frac{5}{6}*\frac{5}{6}\right) +\left( \frac{5}{6}*\frac{1}{6}*\frac{5}{6}\right) +\left(\frac{5}{6}*\frac{5}{6}*\frac{1}{6}\right )$

$=\frac{75}{216}$

Second Case:

Exactly Same Number in next 3 rolls but not 4

$=\frac{5}{6}*\frac{1}{6}*\frac{1}{6} = \frac{5}{216}$

So, Probability of $X$ to  win game is $\frac{75}{216}+ \frac{5}{216} = \frac{80}{216}$

Comments

sir can u elaborate second case??? if no 4 occurs in second case,then x cant win,but the qs demand the prob. of winning...

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3 throws lefts and need 3 exactly same no's

"two dice are already thrown , and having 4, so in remaining three,either we need exactly one 4, or we can have all 1's or all 2's or all 3's or all 5's or all 6's , as in last 3 thrown if we get all similar no but not 4 still X can win"

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oh yes yes,i am contioniously thinkin that if 4 will occur again then x can only win but no any no can turn up thrice..yeah.Thanks..