Answer:
Given, mean = 4, variance = 3 We know that, Mean \[=np\] and variance \[=npq\] \[p+q=1\] and \[p=1-q\] ?(i) According to the question, Mean\[=np\] \[4=np\] ?(ii) and variance\[=npq\] \[3=npq\] \[3=4q\][from Eq. (ii)] \[q=\frac{3}{4}\] On putting the value of q in Eq. (i), we get \[p=1-\frac{3}{4}=\frac{1}{4}\] Now, on putting the value of p in Eq. (ii), we get \[n\times \frac{1}{4}=4\Rightarrow n=16\] Now, binomial distribution \[={}^{n}{{C}_{r}}\cdot {{p}^{r}}{{q}^{n-r}}={}^{16}{{C}_{r}}{{\left( \frac{1}{4} \right)}^{r}}{{\left( \frac{3}{4} \right)}^{16-r}}\]
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