question number 11. the probability that a student correctly answers on the first try (the event a) is…

question number 11. the probability that a student correctly answers on the first try (the event a) is p(a)=0.3. if the student answers incorrectly on the first try, the student is allowed a second try to correctly answer the question (the event b). the probability that the student answers correctly on the second try given that he answered incorrectly on the first try is 0.6. find the probability that the student correctly answers the question on the first or second try. 0.54 0.51 0.18 0.90 0.72 none of the above.

question number 11. the probability that a student correctly answers on the first try (the event a) is p(a)=0.3. if the student answers incorrectly on the first try, the student is allowed a second try to correctly answer the question (the event b). the probability that the student answers correctly on the second try given that he answered incorrectly on the first try is 0.6. find the probability that the student correctly answers the question on the first or second try. 0.54 0.51 0.18 0.90 0.72 none of the above.

Answer

Explanation:

Step1: Identify the probabilities

We know that $P(A) = 0.3$ (probability of correct on first - try), and the probability of correct on second - try given incorrect on first - try is $P(B|\overline{A})=0.6$. The probability of incorrect on first - try is $P(\overline{A})=1 - P(A)=1 - 0.3 = 0.7$.

Step2: Use the law of total probability

The probability of correct on the first or second try is $P(A\cup B)$. Since $A$ and $\overline{A}\cap B$ are mutually - exclusive events, $P(A\cup B)=P(A)+P(\overline{A}\cap B)$. By the formula for conditional probability $P(B|\overline{A})=\frac{P(\overline{A}\cap B)}{P(\overline{A})}$, we can find $P(\overline{A}\cap B)=P(B|\overline{A})\times P(\overline{A})$. So $P(\overline{A}\cap B)=0.6\times0.7 = 0.42$. Then $P(A\cup B)=P(A)+P(\overline{A}\cap B)=0.3 + 0.42=0.72$.

Answer:

0.72