Search any question & find its solution
Question:
Answered & Verified by Expert
If ' \( X \) ' has a binomial distribution with parameters \( n=6, p \) and \( P(X=2)=12, P(X=3)=5 \) then
\( P= \)
Options:
\( P= \)
Solution:
2495 Upvotes
Verified Answer
The correct answer is:
none
Given Options are not matching
$\mathrm{n}=6$
$P(x=r)={ }^{n} C_{r} q^{n-r} p^{r}$
$P(x=2)=12$
${ }^{6} C_{2} q^{4} p^{2}=12$
$P(x=3)=5$
${ }^{6} C_{3} q^{3} p^{3}=5$
$\frac{(1)}{(2)} \Rightarrow \frac{{ }^{6} C_{2} q^{4} p^{2}}{{ }^{6} C_{3} q^{3} p^{3}}=\frac{12}{5}$
$\frac{15 q}{20 p}=\frac{12}{5}$
$75 q=240 p$
$75(1-p)=240 p$
$75-75 p=240 p$
$75=315 p$
$p=\frac{75}{315}=\frac{5}{21}$
$\mathrm{n}=6$
$P(x=r)={ }^{n} C_{r} q^{n-r} p^{r}$
$P(x=2)=12$
${ }^{6} C_{2} q^{4} p^{2}=12$
$P(x=3)=5$
${ }^{6} C_{3} q^{3} p^{3}=5$
$\frac{(1)}{(2)} \Rightarrow \frac{{ }^{6} C_{2} q^{4} p^{2}}{{ }^{6} C_{3} q^{3} p^{3}}=\frac{12}{5}$
$\frac{15 q}{20 p}=\frac{12}{5}$
$75 q=240 p$
$75(1-p)=240 p$
$75-75 p=240 p$
$75=315 p$
$p=\frac{75}{315}=\frac{5}{21}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.