Question

9) A teacher believes one of his students is a "test-dodger." The student missed ten days of a 90-day semester. There were four test days. The student missed three tests. WTP he would miss at least three tests purely by luck? show you work.

Answer 1

Solution:

The given random experiment has two possible outcomes attends the test or misses the test.

The probability of missing the test is:

$\frac{1}{2}=0.5$

We are required to find the probability that a student would miss at least three tests purely by luck.

Let x be the number of tests missed by the student. Therefore, we have to find:

$P(x \geq3)$

We know that the random variable x follows the binomial distribution, therefore using the binomial probability distribution, we have:

$P(x \geq3)=P(x=3)+P(x=4)$

$=\binom{4}{3}0.5^{3}(1-0.5)^{4-3}+\binom{4}{4}0.5^{4}(1-0.5)^{4-4}$

$=(4 \times 0.125\times 0.5)+(1 \times 0.0625 \times 1)$

$=0.25+0.0625$

$=0.3125$

Therefore, the probability he would miss at least three tests purely by luck is 0.3125