Financial Risk Manager Part 1

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Explanation:

The number of breakdowns takes on a binomial distribution with $n = 100$ and $\theta = 0.004$
“Fewer than 3” implies 0, or 1, or 2 machines break down

Therefore,

\begin{aligned} P(\text{fewer than } 3) &= P(0 \text{ breakdowns}) + P(1 \text{ breakdown}) + P(2 \text{ breakdowns}) \\ &= {}_{100}C_0 \cdot 0.004^0 \cdot 0.996^{100} + {}_{100}C_1 \cdot 0.004^1 \cdot 0.996^{99} \\ &\quad + {}_{100}C_2 \cdot 0.004^2 \cdot 0.996^{98} \\ &= 0.6698 + 0.2690 + 0.05347 \\ &= 0.9923 \end{aligned}

Explanation:

The number of breakdowns takes on a binomial distribution with $n = 100$ and $\theta = 0.004$
“Fewer than 3” implies 0, or 1, or 2 machines break down

Therefore,

\begin{aligned} P(\text{fewer than } 3) &= P(0 \text{ breakdowns}) + P(1 \text{ breakdown}) + P(2 \text{ breakdowns}) \\ &= {}_{100}C_0 \cdot 0.004^0 \cdot 0.996^{100} + {}_{100}C_1 \cdot 0.004^1 \cdot 0.996^{99} \\ &\quad + {}_{100}C_2 \cdot 0.004^2 \cdot 0.996^{98} \\ &= 0.6698 + 0.2690 + 0.05347 \\ &= 0.9923 \end{aligned}

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Q.65 A vehicle repairs assembly has a total of 100 jerks and other repair work machines in constant use. The probability of a machine breaking down during a given day is 0.004. There are days when none of the machines break down. However, during some days, one, two, three, four, or more machines break down. Calculate the probability that fewer than 3 machines break down during a particular day.

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RRohit

Last updated: June 22, 2026 at 14:25

0.007726.

0.6698.

0.9923.

0.269.