Question:
If \(X\) follows a Binomial distribution with parameters \(n=8\) and \(p=\frac{1}{2}\) then \(P(|X-4|\leq 2)\)
If \(X\) follows a Binomial distribution with parameters \(n=8\) and \(p=\frac{1}{2}\) then \(P(|X-4|\leq 2)\)
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The event means X lies from 2 to 6, so subtract the probabilities of X equal to 0, 1, 7 and 8 from one.
Updated On: Nov 12, 2026
- \(\frac{59}{64}\)
- \(\frac{119}{128}\)
- \(\frac{117}{128}\)
- \(\frac{29}{32}\)
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The Correct Option is B
Solution and Explanation
Step 1: Understanding the Concept
\(|X-4|\le2\) means \(2\le X\le6\). With \(p=\frac{1}{2}\), \(P(X=r)=\frac{{}^{8}C_{r}}{2^{8}}=\frac{{}^{8}C_{r}}{256}\).
Step 2: Use the complement
\(P=1-[P(0)+P(1)+P(7)+P(8)]\).
Step 3: Compute
\({}^{8}C_{0}=1\), \({}^{8}C_{1}=8\), \({}^{8}C_{7}=8\), \({}^{8}C_{8}=1\), total 18.
So the excluded probability is \(\frac{18}{256}\).
Step 4: Result
\(P=1-\frac{18}{256}=\frac{238}{256}=\frac{119}{128}\).
Final Answer:
The probability is 119/128. \[ \boxed{\text{(B) }\dfrac{119}{128}} \]
\(|X-4|\le2\) means \(2\le X\le6\). With \(p=\frac{1}{2}\), \(P(X=r)=\frac{{}^{8}C_{r}}{2^{8}}=\frac{{}^{8}C_{r}}{256}\).
Step 2: Use the complement
\(P=1-[P(0)+P(1)+P(7)+P(8)]\).
Step 3: Compute
\({}^{8}C_{0}=1\), \({}^{8}C_{1}=8\), \({}^{8}C_{7}=8\), \({}^{8}C_{8}=1\), total 18.
So the excluded probability is \(\frac{18}{256}\).
Step 4: Result
\(P=1-\frac{18}{256}=\frac{238}{256}=\frac{119}{128}\).
Final Answer:
The probability is 119/128. \[ \boxed{\text{(B) }\dfrac{119}{128}} \]
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