Probability

The sample space for the given event is 2 * 2 * 2 = 8. The possible events when tossing a coin three times are, {HTT, TTT, THT, THH, TTH, HHH, HHT, HTH}. Here, there are three events in which even number of heads are there. Hence, the probability will be 3/8.

2 on the red die is divisible by 2 on the green die. 3 on the red die is divisible by 3 on the green die. 5 on the red die is divisible by 5 on the green die. The total number of events is 1 + 1 + 1 = 3 and the sample space is 6 * 6 = 36. Hence, the probability is 3/36.

Total marbles in the bowl = 6 + 5 = 11 The sample space is 11. Now, the probability of getting yellow marbles is 5/11. Hence, the probability of not getting yellow marbles is 1 - 5/11 = 6/11.

Probability for Heads = probability for tails = 1/2 and each trial is independent. Hence required probability= 1/2*1/2*1/2 = 1/8.

Probability of 1st ball red = 6/16 and probability of 2nd blue ball = 7/15

Favourable outcomes are 6 and total are 36. Hence, 1/6 is right.

Right answer; favourable outcomes are 1,2...6. Hence 6/36.

Probability = (5C2)(4C2)/9C4 = 60/126 = 10/21

Either 4 heads and one tail in the fourth or 4 tails and 4 heads. Probability = 2(0.5)^5 = 1/16 = 1/4x which gives x = 4.

The first die can show any numbers from 1 of 6. Second can show numbers greater than the first and third acn show greater than the second. Number of ways = (1,2,3) (1,2,4)(1,2,5)(1,2,6) (1,3,6) (1,4,6) (1,3,5) (1,4,5) (1,4,6)(1,5,6), (2,3,6) (2,3,5)(2,3,4) (2,4,5) (2,4,6) (2,5,6) (3,4,5) (3,4,6) (3,5,6) (4,5,6). Hence, probability = 20/216 = 5/54.

Probability for 5 to be placed in the correct colour = ⁿC_r P^r Q ^{n - r} P = 1/8, Q = 7/8 = 8C5(1/8)^5(7/8)^3 = 1/360

It is independent of what is obtained on first dice. It will be 2/6=1/3

Both cannot be obtained simultaneously.

Favorable outcomes (2,3) (2,6) (4,3) (4,6) (6,6) (3,2) (3,4) (3.6) (6,2) (6,4) total of 11 Favorable outcomes=11/36

3.6 are favorable events. Probability is 1/3