MHT CET · Maths · Probability
Let \(\mathrm{X} \sim \mathrm{B}\left(6, \frac{1}{2}\right)\), then \(\mathrm{P}[|x-4| \leqslant 2]\) is
- A \(\frac{115}{128}\)
- B \(\frac{63}{64}\)
- C \(\frac{57}{64}\)
- D \(\frac{7}{64}\)
Answer & Solution
Correct Answer
(C) \(\frac{57}{64}\)
Step-by-step Solution
Detailed explanation
\(\mathrm{X} \sim \mathrm{~B}\left(6, \frac{1}{2}\right)\)
Here, \(n=6, p=\frac{1}{2}, q=\frac{1}{2}\)
Consider, \(|x-4| \leq 2\)
\(\therefore -2 \leq x-4 \leq 2 \)
\( \therefore 2 \leq x \leq 6 \)
\( \therefore \quad \mathrm{P}[|x-4| \leq 2]=\mathrm{P}(2 \leq x \leq 6) \)
\( =1-[\mathrm{P}(\mathrm{X}=0)+\mathrm{P}(\mathrm{X}=1)] \)
\( =1-\left[{ }^6 \mathrm{C}_0 \mathrm{p}^0 \mathrm{q}^6+{ }^6 \mathrm{C}_1 \mathrm{p}^1 \mathrm{q}^5\right] \)
\( =1-\left[\frac{1}{2^6}+\frac{6}{2^6}\right] \)
\( =1-\frac{7}{64}=\frac{57}{64}\)
Here, \(n=6, p=\frac{1}{2}, q=\frac{1}{2}\)
Consider, \(|x-4| \leq 2\)
\(\therefore -2 \leq x-4 \leq 2 \)
\( \therefore 2 \leq x \leq 6 \)
\( \therefore \quad \mathrm{P}[|x-4| \leq 2]=\mathrm{P}(2 \leq x \leq 6) \)
\( =1-[\mathrm{P}(\mathrm{X}=0)+\mathrm{P}(\mathrm{X}=1)] \)
\( =1-\left[{ }^6 \mathrm{C}_0 \mathrm{p}^0 \mathrm{q}^6+{ }^6 \mathrm{C}_1 \mathrm{p}^1 \mathrm{q}^5\right] \)
\( =1-\left[\frac{1}{2^6}+\frac{6}{2^6}\right] \)
\( =1-\frac{7}{64}=\frac{57}{64}\)
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