MHT CET · Maths · Probability
A die is thrown five times. If getting an odd number is a success, then the probability of getting at least 4 successes is
- A \(\frac{13}{16}\)
- B \(\frac{5}{32}\)
- C \(\frac{1}{32}\)
- D \(\frac{3}{16}\)
Answer & Solution
Correct Answer
(D) \(\frac{3}{16}\)
Step-by-step Solution
Detailed explanation
Here, \(n=5, p=\frac{1}{2}, q=\frac{1}{2}\)
Probability of getting at least 4 successes
\(\mathrm{P}(\mathrm{x}=4 \text { or } \mathrm{x}=5)={ }^5 \mathrm{C}_4\left(\frac{1}{2}\right)^4\left(\frac{1}{2}\right)^1+{ }^5 \mathrm{C}_5\left(\frac{1}{2}\right)^5\left(\frac{1}{2}\right)^{\circ} \)
\( =5 \times \frac{1}{16} \times \frac{1}{2}+1 \times \frac{1}{32} \times 1 \)
\( =\frac{5}{32}+\frac{1}{32} \)
\( =\frac{6}{32}=\frac{3}{16}\)
Probability of getting at least 4 successes
\(\mathrm{P}(\mathrm{x}=4 \text { or } \mathrm{x}=5)={ }^5 \mathrm{C}_4\left(\frac{1}{2}\right)^4\left(\frac{1}{2}\right)^1+{ }^5 \mathrm{C}_5\left(\frac{1}{2}\right)^5\left(\frac{1}{2}\right)^{\circ} \)
\( =5 \times \frac{1}{16} \times \frac{1}{2}+1 \times \frac{1}{32} \times 1 \)
\( =\frac{5}{32}+\frac{1}{32} \)
\( =\frac{6}{32}=\frac{3}{16}\)
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