Go To Your Study Dashboard

Get Answers to all your Questions

header-bg

Let a random variable X have a binomial distribution with mean $8$ and variance $4$ . If $P\left ( X\leq 2 \right )=\frac{k}{2^{16}}$ , then k is equal to :

Option 1)
$137$

Option 2)
$17$

Option 3)
$121$

Option 4)
$1$

Answers (1)

best_answer

$np=8$

$npq=4$

$q=\frac{1}{2}\: \: \: \: \: ,\: \: \: n=16$

$\Rightarrow p=\frac{1}{2}$

$P\left ( x=r \right )=^{16}C{_{r}}\left ( \frac{1}{2} \right )^{16}$

$P\left ( x\leq 2 \right )=\frac{16C_{0}+16C_{1}+16C_{2}}{2^{16}}$

$=\frac{137}{2^{16}}$

So, the value of $k=137$

Option 1)

$137$

Option 2)

$17$

Option 3)

$121$

Option 4)

$1$

Posted by

Plabita

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

JEE Main 2026 correction window opens tomorrow; NTA allows exam city change

anam-khan

Faridabad: 17-year-old girl preparing for JEE Main 2026 dies by suicide

anam-khan

NTA JEE Mains 2026 application closes at 9 am today; correction facility from December 1

Latest Question