Final answer:
To find the probability of being selected in at least one firm, we calculated the probability of being selected in both and subtracted this from the complement of both rejections. The final total probability of being selected in at least one firm is 0.4. Hence, the correct option is B. 0.4.
Step-by-step explanation:
The student's question revolves around calculating the probability of being selected in at least one of two firms, given certain probabilities for selection and rejection. We know the probability of being selected in firm X is 0.7 and the probability of being rejected at firm Y is 0.5. To find the probability of being selected in one of the firms, we must consider the given probability of at least one rejection, which is 0.6.
The probability of being selected in both firms would be the product of the probabilities of being selected in each, due to independence: P(X and Y) = P(X) * P(Y). Since we know P(X) and P(rejected at Y), we can find P(Y) by subtracting the probability of rejection at Y from 1, so P(Y) = 1 - 0.5 = 0.5.
The probability of being selected in both is then P(X and Y) = 0.7 * 0.5 = 0.35. Now, we can find the probability of being selected in at least one firm, which is the complement of being rejected in both. Since the probability of at least one rejection is 0.6, the probability of no rejections (being selected in both) is 1 - 0.6 = 0.4. However, this includes the probability of being selected in both firms, which we calculated as 0.35. So, the probability of being selected in exactly one firm is 0.4 minus the probability of being selected in both firms, which gives us 0.4 - 0.35 = 0.05.
Therefore, the total probability of being selected in at least one of the firms is the sum of the probabilities of being selected in exactly one firm and in both firms, which is 0.05 + 0.35 = 0.4. Hence, the correct option is B. 0.4.