In a college, it has been found that 40% of the students withdraw without completing the mathematics course if there were 10 students have registered for the course of this semester. a) What is the…

Solution

This problem follows a binomial probability distribution where:

• Number of trials (n) = 10 (students registered)
• Probability of success (p) = Probability of withdrawal = 40% = 0.4
• Probability of failure (q) = Probability of not withdrawing = 60% = 0.6

The binomial probability formula is: $$P(X=k)=\left(\genfrac{}{}{0ex}{}{n}{k}\right)p^k{q}^{n-k}$$

a) Probability that none will withdraw

Here, $k=0$ (no withdrawals). $$P(X=0)=\left(\genfrac{}{}{0ex}{}{10}{0}\right)(0.4{)}^0(0.6{)}^{10}$$ $$=1\times 1\times (0.6{)}^{10}$$ $$=(0.6{)}^{10}$$ $$=0.0060466176\approx 0.00605$$

Answer: The probability that none will withdraw is 0.00605 (or 0.605%).

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b) Probability that at least one will withdraw

This is the complement of "none withdraw": $$P(X\ge 1)=1-P(X=0)$$ $$=1-0.00605$$ $$=0.99395$$

Answer: The probability that at least one will withdraw is 0.99395 (or 99.395%).

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c) Probability that at most one will withdraw

This includes two cases: 0 withdrawals and 1 withdrawal. $$P(X\le 1)=P(X=0)+P(X=1)$$

1. For $k=0$: $$P(X=0)=0.00605\text{(from part a)}$$

2. For $k=1$: $$P(X=1)=\left(\genfrac{}{}{0ex}{}{10}{1}\right)(0.4{)}^1(0.6{)}^9$$ $$=10\times 0.4\times (0.6{)}^9$$ $$=4\times 0.010077696$$ $$=0.040310784\approx 0.0403$$

3. Total probability: $$P(X\le 1)=0.00605+0.0403$$ $$=0.04635$$

Answer: The probability that at most one will withdraw is 0.04635 (or 4.635%).

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