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Exercise 12.5 · Q3

Q.AA, BB, and CC try to hit a target simultaneously but independently. Their respective probabilities of hitting the target are 34,12,58\dfrac{3}{4}, \dfrac{1}{2}, \dfrac{5}{8}. The probability that the target is hit by AA or BB but not by CC is

(1) 2164\dfrac{21}{64}
(2) 732\dfrac{7}{32}
(3) 964\dfrac{9}{64}
(4) 78\dfrac{7}{8}
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✓ Free question

Step 1. Given. P(A)=34P(A)=\dfrac34, P(B)=12P(B)=\dfrac12, P(C)=58P(C)=\dfrac58, all independent.

Step 2. Find P(A∪B)P(A\cup B). P(A∪B)=P(A)+P(B)−P(A)P(B)=34+12−34×12=34+12−38=68+48−38=78P(A\cup B)=P(A)+P(B)-P(A)P(B)=\dfrac34+\dfrac12-\dfrac34\times\dfrac12=\dfrac34+\dfrac12-\dfrac38=\dfrac68+\dfrac48-\dfrac38=\dfrac78.

Step 3. Find P(Cˉ)P(\bar C). P(Cˉ)=1−58=38P(\bar C)=1-\dfrac58=\dfrac38.

Step 4. Combine (independence of CC from the others). P((A∪B)∩Cˉ)=P(A∪B)×P(Cˉ)=78×38=2164P((A\cup B)\cap\bar C)=P(A\cup B)\times P(\bar C)=\dfrac78\times\dfrac38=\dfrac{21}{64}.

✓Final answer

P=2164P=\dfrac{21}{64} -- option (1).

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