A pair of dice is thrown 7 times. If 'getting a total of $7^{\prime}$ is considered a success, find the probability of getting(i) no success(ii) exactly 6 successes

Home » A pair of dice is thrown 7 times. If ‘getting a total of is considered a success, find the probability of getting(i) no success(ii) exactly 6 successes

A pair of dice is thrown 7 times. If ‘getting a total of $7^{\prime}$ is considered a success, find the probability of getting
(i) no success
(ii) exactly 6 successes

A pair of dice is thrown 7 times. If ‘getting a total of $7^{\prime}$ is considered a success, find the probability of getting
(i) no success
(ii) exactly 6 successes

(i) Using Bernoulli’s Trial $P($ Success $=x)={ }^{n} C_{x} \cdot p^{x} \cdot q^{(n-x)}$
$x=0,1,2, \ldots \ldots . . n$ and $q=(1-p), n=7$
the favourable outcomes,
$(1,6),(6,1),(2,5),(5,2),(3,4),(4,3)$
The probability of success $=p=\frac{6}{36}=\frac{1}{6}$
$\mathrm{q}=\frac{5}{6}$
probability of no success $\left.={ }^{7} C_{0} \cdot \frac{(1}{6}\right)^{0}\left(\frac{5}{6}\right)^{7}$
$\Rightarrow\left(\frac{5}{6}\right)^{7}$
(ii) Using Bernoulli’s Trial $P($ Success $=x)={ }^{n} C_{x} \cdot p^{x} \cdot q^{(n-x)}$ $x=0,1,2, \ldots \ldots . . n$ and $q=(1-p), n=7$
the favourable outcomes,
$(1,6),(6,1),(2,5),(5,2),(3,4),(4,3)$
The probability of success $=p=\frac{6}{36}=\frac{1}{6}$
$\mathrm{q}=\frac{5}{6}$
probability of exactly 6 successes $={ }^{7} C_{6} \cdot\left(\frac{1}{6}\right)^{6}\left(\frac{5}{6}\right)^{1}$
$\Rightarrow 35 .\left(\frac{1}{6}\right)^{7}$