A and B aim a target. The Probability that A hits the target is 5/7 and B hits the target in 7/10. What is the probability that exactly one of them hits the target?

Required probability = Probability of getting 2 tails + Probability of getting 1 tail + Probability of getting no tail.
= ⁸C₂ x 1/256 + ⁸C₁ x 1/256 + ⁸C₀ x 1/256
= 37/256

n(S) = 6 x 6 = 36
n(E) = (1, 2) (2, 1), (1, 5) (2, 4) (3, 3) (4, 2) (5, 1) (3, 6) (4, 5) (5, 4) (6, 3) (6, 6) = 12
? P(E) = 12/36

Total number of ways of selecting 1 card from 200 cards is ²⁰⁰C₁ ways.
Let E be an event of drawing a number which is a perfect cube.
E = {1, 8, 27, 64, 125 }
? P(E) = n(E) / n(S)
= ⁵C₁ / ²⁰⁰C₁ = 5/200 = 1/40

Probability of head or a tail on upper side of coin = 1/2
? Probability of getting same face on all three coins = 1/2 x 1/2 x 1/2
= 1/8

Total number of cards = 104
Total number of jacks = 8
? Probability for the jack in first draw = 8/104
and probability for the jack in second draw = 7/103
Since, both are independence events.
? Required probability = 8/104 x 7/103 = 7/1339

Probability that first ball drawn is white = ⁵C₁ = 1/4
Since, balls are drawn with replacement, hence all the four events will have equal probability.
Therefore, required probability = 1/4 x 1/4 x 1/4 x 1/4 = 1/256

Total number of balls in the bag = 12
Probability of drawing one blue ball in the first draw P(E₁) = ⁸C₁ / ¹²C₁
= 8/12 = 2/3

After the first drawn of a blue ball, now there are 7 blue and 4 white ball in the bag. Total number of the balls in the bag is 11.

Probability of drawing one blue ball in the second drawn = P(E₂) = ⁷C₁ /¹¹C₁ = 7/11

? Probability that both are blue P(E₁ ? E₂) = P(E₁) x P(E₂)
= 2/3 x 7/11 = 14 / 33

Here, n(S) = 6 x 6 = 36 and E = Event of getting a number other than 4 on any dice
= {(1, 1), (1, 2), (1, 3), (1, 5), (1, 6), (2, 1), (2, 2) , (2, 3) , (2, 5), (2, 6), (3, 1), (3, 2), (3, 3), (3, 5), (3, 6), (5, 1) , (5, 2) , (5, 3), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 5), (6, 6) }
? P(E) = n(E) / n(S) = 25/36

Total number of ways of selecting one number from 25 number is ²⁵C₁ ways.
Let E be the event of drawing a number divisible by both 2 and 3
i.e., (if a number is divisible by both 2 and 3, it is divisible by 6 also)
E = {6, 12, 18, 24}
? Required probability = P(E) = n(E) / n(S) = 4/25

Time taken for complete cycle = 60 s
Probability that light will be green = (Time for which light is green)/(time taken for complete cycle)
= 25/60 = 5/12
? Probability that light will not be green = 1- 5/12 = 7/12