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Q. 12 Find the probability of throwing at most $2$ sixes in $6$ throws of a single die.

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Let X represent number of times getting 2 six in 6 throws of a die.

Probability of getting 6 in single throw of die=P

$P=\frac{1}{6}$

$q=1-P=1-\frac{1}{6}=\frac{5}{6}$

X has a binomial distribution,n=6

$\therefore \, \, \, \, P(X=x)=^nC_x.q^{n-x}.p^x$

$P(X=x)=^6C_x.$ $(\frac{5}{6})^{6-x} . (\frac{1}{6})^{x}$

$P(getting\, \, atmost\, \, two\, six)=P(X\leq 2)$

$=P(X=0)+P(X=1)+P(X=2)$

$=^{6}C_0 (\frac{5}{6})^{6}\frac{1}{6}^0+^{6}C_1 (\frac{5}{6})^{5}\frac{1}{6}^1+^{6}C_2 (\frac{5}{6})^{4}\frac{1}{6}^2$

$=1.(\frac{5}{6})^{6}+6. (\frac{5}{6})^{5}\frac{1}{6}+15 (\frac{5}{6})^{4}\frac{1}{36}$

$=(\frac{5}{6})^{6}+ (\frac{5}{6})^{5}+ (\frac{5}{6})^{4}\frac{5}{12}$

$=(\frac{5}{6})^{4}( \frac{5}{6}^{2}+ \frac{5}{6}+\frac{5}{12})$

$=(\frac{5}{6})^{4}( \frac{25}{36}+ \frac{5}{6}+\frac{5}{12})$

$=(\frac{5}{6})^{4}( \frac{70}{36})$

$=(\frac{5}{6})^{4}( \frac{35}{18})$

Posted by

seema garhwal

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