For what choice of p are a and b independent solution

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For what choice of P are A and B independent Solution We have P (A or B)= P (A) + P (B) P (AB) P (AB)=P (A) + P(B) P (A or B) =P + 0.4 0.8 =P 0.4 If A and B are mutually exclusive P (A or B)=0 O=P 0.4 P= 0.4 A and B are independent if P (AB)=P (A) P (B) P 0.4=0.4 P (1 0.4) P=0.4 0.6 P=0.4 p= 0.4/0.6 =0.67 Illustration 7 A person is known to hit the target in 3 out of 4 shots, whereas another person is known to hit the target in 2 out of 3 shots. Find the probability of the target being hit at all when the both try.
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14 Solution Probability of 1st person hits the target (A) = 3/4 Probability of 2 nd person hits the target (B) = 2/3 The events are not mutually exclusive because both of them try of hit the target. So, the addition theorem is used. P (A or B)= P (A) + P (B) P (AB) = 3 2 4 3 3 2 4 3 = 917 . 0 12 11 12 6 12 17 Illustration 8 a. A product is assembled from three components, X, Y, and Z, the probability of these components being defective is respectively 0.01, 0.02 and 0.05. What is the probability that the assembled product will not be defective. b. Items produced by a certain process each may have one or both the two types of defects, A and B. It is known that 20 per cent of the items have type A defects and 10 per cent have type B defects. Furthermore, 6 per cent are known to have both types of defects. What is the probability that a randomly selected item will be defective. Solution a. Denote, A, B and C is components of X, Y and Z being defective respectively. We are given: P (A) = 0.01, P (B) = 0.02, P (C) = 0.05 Probability that the assembled product will be defective. P A or B or C) = P (A) + P (B) + P (C) P (AB) P (BC) P (AC) + P (ABC) = 0.01 + 0.02 + 0.05 - 0.01 0.02 - 0.02 0.05 - -0.01 0.05 + 0.01 0.02 0.05 = 0.08 - 0.0017 + 0.00001 = 0.07831 Hence, required probability that assembled product not be defective = 1 0.07831 = 0.92169 or 0.922. b. We are given 20 . 0 100 20 P(A) 10 . 0 100 10 P(B)
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15 06 . 0 100 6 P(AB) Probability of randomly selected item will be defective P (A or B)= P (A) + P (B) P (AB) = 0.20 + 0.10 0.06 = 0.24 Illustration 9 A salesman has 65 per cent chance of making a sale to each customer. The behaviour of successive customer is independent. If three customers P, Q and R enter, what is the probability that salesman will make a sale to A or B or C. Solution In notation A = Probability of making sale to P B = Probability of making sale to Q C = Probability of making sale to R P (A)= 0.65, P (B) = 0.65, P(C) = 0.65 P (A) or (B) or (C) = P (A) + P (B) + P (C) P (AB) P (BC) P (AC) + P (ABC) = 0.65 + 0.65 + 0.65 0.65 0.65 0.65 0.65 0.65 0.65 + 0.65 0.65 0.65 = 1.95 1.2675 + 0.2746 = 0.9571 3.3.3. MULTIPLICATION THEOREM The probability of occurring of two independent events A and B is equal to the product of their individual probabilities. Symbolically, if A and B are independent, then P (A and B)= P (A) P (B) Proof If an event a can happen in n 1 ways of which a 1 are successful, we can combine each successful event in the first with each successful event in the second.
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