Probability Question 13
Question 13 - 01 February - Shift 1
рдПрдХ рдмрд╛рдИрдиреЛрдорд┐рдпрд▓ рд╡рд┐рддрд░рдг $B(n, p)$ рдореЗрдВ, рдорд╛рдзреНрдп рдФрд░ рд╡рд┐рдЪрд▓рди рдХреЗ рдпреЛрдЧ рдФрд░ рдЧреБрдгрдирдлрд▓ рдХреНрд░рдорд╢рдГ 5 рдФрд░ 6 рд╣реИрдВ, рддреЛ $6(n+p-q)$ рдХрд┐рд╕рдХреЗ рдмрд░рд╛рдмрд░ рд╣реИ:-
(1) 51
(2) 52
(3) 53
(4) 50
Show Answer
Answer: (2)
Solution:
Formula: Mean and variance of probability distribution , Probability of an event , Roots of equations , Important results on probability
$n p+n p q=5, n p \cdot n p q=6$
$np(1+q)=5, n^{2} p^{2} q=6$
$n^{2} p^{2}(1+q)^{2}=25, n^{2} p^{2} q=6$
$\frac{6}{q}(1+q)^{2}=25$
$6 q^{2}+12 q+6=25 q$
$6 q^{2}-13 q+6=0$
$6 q^{2}-9 q-4 q+6=0$
$(3 q-2)(2 q-3)=0$
$q=\frac{2}{3}, \frac{3}{2}, q=\frac{2}{3}$ is accepted
$p=\frac{1}{3} \Rightarrow$ n. $\frac{1}{3}+n \cdot \frac{1}{3} \cdot \frac{2}{3}=5$
$\frac{3 n+2 n}{9}=5$
$n=9$
So $6(n+p-q)=6(9+\frac{1}{3}-\frac{2}{3})=52$
рдмреАрдЯрд╛
