MHT CET · Maths · Probability
The probability that bomb will miss the target is \(0 \cdot 2\). Then the probability that out of 10 bombs dropped exactly 2 will hit the target is
- A \(\frac{288}{5^{10}}\)
- B \(\frac{144}{5^{9}}\)
- C \(\frac{144}{5^{10}}\)
- D \(\frac{288}{5^{9}}\)
Answer & Solution
Correct Answer
(B) \(\frac{144}{5^{9}}\)
Step-by-step Solution
Detailed explanation
\((\mathrm{C})\)
We have probability of missing the target \(0.2\)
Let \(\mathrm{p}=1-0.2=0.8, \mathrm{q}=0.2, \mathrm{n}=10\) and \(\mathrm{r}=2\)
Hence required probability \(={ }^{10} \mathrm{C}_{2}(0.8)^{2}(0.2)^{8}\)
\(=\frac{10 !}{2 ! 8 !} \times\left(\frac{8}{10}\right)^{2} \times\left(\frac{2}{10}\right)^{8}=\frac{45 \times 64 \times 2^{8}}{10^{10}}=\frac{9 \times 5 \times 2^{14}}{2^{10} \times 5^{10}}=\frac{9 \times 2^{4}}{5^{9}}\) \(=\frac{144}{5^{9}}\)
We have probability of missing the target \(0.2\)
Let \(\mathrm{p}=1-0.2=0.8, \mathrm{q}=0.2, \mathrm{n}=10\) and \(\mathrm{r}=2\)
Hence required probability \(={ }^{10} \mathrm{C}_{2}(0.8)^{2}(0.2)^{8}\)
\(=\frac{10 !}{2 ! 8 !} \times\left(\frac{8}{10}\right)^{2} \times\left(\frac{2}{10}\right)^{8}=\frac{45 \times 64 \times 2^{8}}{10^{10}}=\frac{9 \times 5 \times 2^{14}}{2^{10} \times 5^{10}}=\frac{9 \times 2^{4}}{5^{9}}\) \(=\frac{144}{5^{9}}\)
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