If X sim Bleft(8, frac{1}{2}right),then P(|X- | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Given $X \sim B\left(8, \frac{1}{2}\right)$,we have $n=8, p=\frac{1}{2}, q=\frac{1}{2}$.We need to find $P(|X-4| \leq 2)$.This inequality $|X-4| \

Vedclass · Accepted Answer

$\frac{119}{128}$

Vedclass · Answer

(A) Given $X \sim B\left(8, \frac{1}{2}\right)$,we have $n=8, p=\frac{1}{2}, q=\frac{1}{2}$.We need to find $P(|X-4| \leq 2)$.This inequality $|X-4| \leq 2$ implies $-2 \leq X-4 \leq 2$,which simplifies to $2 \leq X \leq 6$.Thus,we need to calculate $P(X=2) + P(X=3) + P(X=4) + P(X=5) + P(X=6)$.The probability mass function is $P(X=k) = {}^{8}C_{k} \left(\frac{1}{2}\right)^{k} \left(\frac{1}{2}\right)^{8-k} = {}^{8}C_{k} \left(\frac{1}{2}\right)^{8}$.Summing these probabilities:$P(2 \leq X \leq 6) = \left(\frac{1}{2}\right)^{8} \left[ {}^{8}C_{2} + {}^{8}C_{3} + {}^{8}C_{4} + {}^{8}C_{5} + {}^{8}C_{6} \right]$.Calculating the combinations:${}^{8}C_{2} = 28, {}^{8}C_{3} = 56, {}^{8}C_{4} = 70, {}^{8}C_{5} = 56, {}^{8}C_{6} = 28$.Sum $= 28 + 56 + 70 + 56 + 28 = 238$.Therefore,$P(2 \leq X \leq 6) = \frac{238}{256} = \frac{119}{128}$.

If $X \sim B\left(8, \frac{1}{2}\right)$,then $P(|X-4| \leq 2) = $

Similar Questions

It is observed that there will be $25$ blood specimens of normal persons,if $100$ blood samples are tested. If $10$ specimens are sent to a laboratory for testing,then the probability of having at least two specimens of normal persons is

Two cards are drawn successively with replacement from a well-shuffled pack of $52$ cards. Let $X$ denote the random variable of the number of kings obtained in the two drawn cards. Then $P(X=1) + P(X=2)$ equals:

Find the probability distribution of the number of doublets in three throws of a pair of dice.

The probability of guessing correctly at least $7$ out of $10$ answers in a "True" or "False" test is = $ . . . . . . $

As a business strategy, $20 \%$ of the new internet service subscribers selected randomly receive a special promotion. If a group of $5$ such subscribers signs for the service, then the probability that at least two of them get the special promotion is

Vedclass Test Series

Exam Paper Generator

Online Exam Module