Probability Distributions

The word “MUMBAI” has six letters. Therefore, total ways of arrangements = 6!/2 =360. Number of words where two M’s do not come together is = Total words – Number of words where they come together = 360 – 5! = 240 Hence, the required probability is = 240/360 = 2/3.

Number of ways in which six letters can be arranged is 6!/2 = 720/2 = 360 ways. If N always comes next to A, they both can be treated as a single letter. In that case, the total arrangements will be, 5!. Therefore, the required probability is 5! / 360 = 1/3.

Let, A = probability of first event taking place, P(A) = 3/(7 + 3) = 3/10. B = probability of second event taking place, P(B) = 5/(5 + 2) = 5/7. The event will occur if  either only A takes place and B does not take place or B takes place and A does not take place, or both the events take place. Therefore, the required probability will be = [P(A) x {1 - P(B)}] + [P(B) x {1 - P(A)}] + [P(A) x P(B)] = [3/10 x (1 - 5/7)] + [5/7 x (1 - 3/10)] + [3/10 x 5/7] = 4/5.

Probability that John can solve the problem is 1/3. Probability that John cannot solve the problem is (1 - 1/3) = 2/3. Probability that Johny can solve the problem is 1/5. Probability that Johny cannot solve the problem is (1-1/5) = 4/5. Probability that Janet can solve the problem is 1/6. Probability that Janet cannot solve the problem is (1 - 1/6) = 5/6. The probability that they together cannot solve the problem is = 2/3 × 4/5 × 5/6 = 4/9. Therefore, the required probability that the problem will be solved is = (1 - 4/9) = 5/9.

Let, P(R) = Probability of John’s selection = 1/5.  So, the probability that John is not selected is P(R ̅) = 1 - 1/5 = 4/5. Similarly, P(A) = Probability of Janet’s selection = 1/7. So, P(A ̅) = 1 - 1/7 = 6/7. Hence, the probability that only one of them will be selected is = P(R) × P(A ̅) + P(A) × P(R ̅) = 2/7.

Let R: John speaks the truth; and A: Janet speaks the truth. Hence, P(R) = 70/100 and P (R ̅) = 1 - 70/100 = 30/100. Again, P (A) = 80 /100 and P (A ̅) = 1 - 80/100 = 20/100. Therefore, required probability,P(R) × P (A ̅) + P (A) × P (R ̅) = 70/100 x 20/100 + 80/100 x 30/100 = 19/ 50

There are 5 letters. The probability, that the letter I gets the first position is 1/5. The probability that M is in the second position is 1/4. Similarly, the probability for A, G and E are 1/3 , 1/2, 1/1,  respectively. Hence, the required probability is = (1/5 × 1/4 × 1/3 × 1/2 × 1/1 ) = 1/120.

The probability that the system will not fail for 4 days is 0.0625.

Chances that exactly 2 of them will be children = Favourable outcomes/Total outcomes = 10/21. Option 1 is correct.

Probability that none of the men and women will get selected = (1 - 1/4 )x (1 - 1/3) = 3/4 x 2/3 = 1/2 Option 1 is correct.

Odd against event E1 = 5/2 Probability of event E1 = 2/7 Odds in favour of event E2 = 6/5 Probability of event E2 = 6/11 Probability that none of the events will occur = (1 - 2/7 ) (1 - 6/11) = 5/7 x 5/11 = 25/77 Option 4 is correct.

Using binomial distribution with n=5 and p=0.5, the probability of at most one negative value is P(X=0) + P(X=1). P(X=0) = (0.5)^5 = 1/32. P(X=1) = 5 * (0.5)^1 * (0.5)^4 = 5/32. Total = 6/32.

Using Bayes' Theorem: P(Y|Reliable) = P(Reliable|Y) * P(Y) / [P(Reliable|Y) * P(Y) + P(Reliable|X) * P(X)]. P(Y|Reliable) = (0.72 * 0.40) / (0.72 * 0.40 + 0.96 * 0.60) = 0.288 / (0.288 + 0.576) = 0.288 / 0.864 = 0.3333.

Probability of car = 0.45. Probability of public transport = 1 - 0.45 = 0.55. Within public transport, bus = 0.55 * 0.55 = 0.3025 (approx 0.30) and metro = 0.55 * (1 - 0.55) = 0.55 * 0.45 = 0.2475 (approx 0.25).

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