Two-way tables give intersection probabilities | My Assignment Tutor

1. Two-way tables give intersection probabilities, and you are asked for conditional probabilities. You needthe formulaP(AjB) = P(A B)P(B) :(a) 108 329 ≈ 0:33(b) 108 371 ≈ 0:292. This is a binomial distribution with p = 0:5.(a) P(X = 2) = 3 20:520:53-2 = 0:375(b) At least one tail” can happen in 7 ways:fHHT; HT H; T HH; HT T; T HT; T T H; T T T g:Of those ways (reducing your sample space to the given event), 4 of them have at least 2 tails. Therefore,P(X ≥ 2jX ≥ 1) = 4 7 ≈ 0:57.3. Define the eventsA: the product breaks down in the third year;B: the product lasts at least 2 years.Then we want to find P(AjB), which equals PP(A(BB) ).P(B) = P(T ≥ 2) = e- 2 5P(A) = P(2 ≤ T ≤ 3) = P(T ≥ 2) – P(T ≥ 3)= e-25– e-35Since A ⊂ B, we have A B = A. Therefore,P(AjB) = P(A B)P(B)=P(A)P(B)=e-25– e-35e-25≈ 0:18134.(a)P(X = 0) = 5 00:700:35 = 0:00243P(X = 1) = 5 10:710:34 = 0:02835P(X = 2) = 5 20:720:33 = 0:1323P(X = 3) = 5 30:730:32 = 0:3087P(X = 4) = 5 40:740:31 = 0:36015P(X = 5) = 5 50:750:30 = 0:168070 1 2 3 4 50.00 0.10 0.20 0.30xy0 1 2 3 4 50.0 0.2 0.4 0.6 0.8 1.0xcumulative(b) The others will probably score less than your most likely score”, which is 4. So you will probably need tomake 4 out of 5 baskets. P(x ≥ 4) = 0:36015 + 0:16807 = 0:528225..Notice that 0:008 + 0:392 + 0:024 + 0:576 = 1. We have P(D) = P(Y D) + P(O D) = 0:008 + 0:024 = 0:032Now we have a binomial situation with P(D) = 0:032 and P(G) = 0:968. The probability that at most 2 aredefective isP(X ≤ 2) = P(X = 0) + P(X = 1) + P(X = 2)= 30 0 0:03200:96830 + 30 1 0:03210:96829 + 30 2 0:03220:96828≈ 0:92

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