Let W stand for the winning of a game and L for losing it. Then there are 4 mutually exclusive possibilities

(i) W, W, W
(ii) W, W, L, W
(iii) W, L, W, W
(iv) L, W, W, W.

[Note that case (i) includes both the cases whether he losses or wins the fourth game.]

By the given conditions of the question, the probabilities for (i), (ii), (iii) and (iv) respectively are

Hence the required probability

= 

[Since the probability of winning the game if previous game was also won is 2/(1+2) = 2/3 and the probability of winning the game if previous game was a loss is 1/(1+2) = 1/3].

Probability that only husband is selected

Total of seven can be obtained in the following ways

1, 1, 1, 4 in = 4 ways [there are four objects, three repeated]

Similarly,

1, 1,2, 3 in = 12 ways

1, 2,2, 2 in  = 4 ways

Hence, required probability

[Since exhaustive no. of cases = 6 × 6 × 6 × 6 = 6⁴]

The number of ways of getting the different number 1, 2, ….., 6 in six dice = 6!.

Total number of ways = 6⁶

Hence, required probability

= 

The faulty machines can be identified in two tests only if both the tested machines are either all defective or all non-defective. See the following tree diagram.

(Here D is for Defective & ND is for Non Defective)

Required Probability

= 

Since the probability that first machine is defective (or non-defective) is ²⁄₄ and the probability that second machine is also defective (or non – defective) is ⅓ as 1 defective machine remains in total three machines.

The probability that the person hits the target = 0.3

∴ The probability that he does not hit the target in a trial

= 1 – 0.3 = 0.7

∴ The probability that he does not hit the target in any of the ten trials = (0.7)¹⁰

∴ Probability that he hits the target

= Probability that at least one of the trials succeeds

= 1 – (0.7)¹⁰.

The probability that A cannot solve the problem

Seven people can seat themselves at a round table in 6! ways. The number of ways in which two distinguished persons will be next to each other = 2 (5) !, Hence, the required probability

The condition implies that the last digit in both the integers should be 0, 1, 5 or 6 and the probability

[ Since the squares of numbers ending in 0 or 1 or 5 or 6 also 0 or 1 or 5 or 6 respectively]

Exhaustive number of cases = 12

Favourable cases = ¹²C₂ (2⁶ – 2)

∴ Probability