The probability of student A passing an exam is 2/7 and that of B passing is 5/7. If these probabilities are independent, what is the probability that only B passes the examination
25/49
Understanding the probabilities of independent events is key to solving this problem. We are given the individual probabilities of student A and student B passing an exam, and we need to find the probability that only student B passes the examination.
Two events are considered independent if the outcome of one does not affect the outcome of the other. In this question, the passing of student A and student B are independent events. This means we can multiply their individual probabilities to find the probability of both occurring (or one occurring and the other not).
To find the probability that only B passes, we must also consider the event that A fails. The probability of an event not occurring is 1 minus the probability of it occurring.
Probability of student A failing the exam, denoted as \(P(A')\):
\(P(A') = 1 - P(A)\)
\(P(A') = 1 - \frac{2}{7}\)
\(P(A') = \frac{7 - 2}{7}\)
\(P(A') = \frac{5}{7}\)
The event "only B passes the examination" means that student B passes AND student A fails. Since these are independent events, we can multiply their probabilities:
Probability (only B passes) = \(P(\text{B passes and A fails})\)
Probability (only B passes) = \(P(B) \times P(A')\)
Substitute the calculated values:
Probability (only B passes) = \(\frac{5}{7} \times \frac{5}{7}\)
Probability (only B passes) = \(\frac{5 \times 5}{7 \times 7}\)
Probability (only B passes) = \(\frac{25}{49}\)
The probability that only student B passes the examination is \(\frac{25}{49}\). This calculation clearly demonstrates the application of independent probability rules for exam scenarios.
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