MHT CET · Maths · Probability
Out of 100 people selected at random, 10 have common cold. If five persons selected at random from the group, then the probability that at most one person will have common cold is
- A \(0.9254\)
- B \(0.9185\)
- C \(0.9851\)
- D \(0.9245\)
Answer & Solution
Correct Answer
(B) \(0.9185\)
Step-by-step Solution
Detailed explanation
(C)
Let probability of having common cold \(\mathrm{p}=\frac{10}{100}=\frac{1}{10} \Rightarrow \mathrm{q}=\frac{9}{10}\) Here \(\mathrm{n}=5\), and \(\mathrm{x}=0,1\).
Hence required probability
\(\begin{array}{l}
={ }^{5} \mathrm{C}_{0}\left(\frac{1}{10}\right)^{0}\left(\frac{9}{10}\right)^{5}+{ }^{5} \mathrm{C}_{1}\left(\frac{1}{10}\right)^{1}\left(\frac{9}{10}\right)^{4} \\
=\left(\frac{9}{10}\right)^{5}+\frac{5(9)^{4}}{(10)^{5}}=\frac{\left(9^{4}\right)[9+5]}{(10)^{5}}=\frac{81 \times 81 \times 14}{10^{5}}=0.9185
\end{array}\)
Let probability of having common cold \(\mathrm{p}=\frac{10}{100}=\frac{1}{10} \Rightarrow \mathrm{q}=\frac{9}{10}\) Here \(\mathrm{n}=5\), and \(\mathrm{x}=0,1\).
Hence required probability
\(\begin{array}{l}
={ }^{5} \mathrm{C}_{0}\left(\frac{1}{10}\right)^{0}\left(\frac{9}{10}\right)^{5}+{ }^{5} \mathrm{C}_{1}\left(\frac{1}{10}\right)^{1}\left(\frac{9}{10}\right)^{4} \\
=\left(\frac{9}{10}\right)^{5}+\frac{5(9)^{4}}{(10)^{5}}=\frac{\left(9^{4}\right)[9+5]}{(10)^{5}}=\frac{81 \times 81 \times 14}{10^{5}}=0.9185
\end{array}\)
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