Mathematics
alinaqureshi
2015-11-19 06:23:12
Guys i need help on this one pliz help me out
ANSWERS
KaneshaNorth
2015-11-19 07:18:25

The probability of r successes in n trials is [latex]C(n,r)p^r(1-p)^{n-r}[/latex]  where C(n,r) is the number of combinations of n things taken r at a time... [latex]C(n,r)=frac{n!}{r!(n-r)!}[/latex] Define "success" as getting a 6 on the die.  Then the probability of success is [latex]p=frac{1}{6} [/latex] and the probability of "failure" is [latex]1-p=frac{5}{6} [/latex]. The probility is [latex]C(4,2)left(frac{1}{6} ight)^2 left(frac{5}{6} ight)^2=6left(frac{1}{36} ight) left( frac{25}{36} ight)=0.11574=11.6%[/latex]

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