Important Questions

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Probability

13.1. Conditional Probability.

Q.1. A and B toss a coin alternately till one of them gets a head and wins the game. If A starts first, find the probability that B will win the game.

Solution :

We have, P(H) = 1/2 and P(T) = 1/2 .
Therefore, P(B wins the game) = P(TH or TTTH or TTTTTH or -----)
= P(TH) + P(TTTH) + P(TTTTTH)
= P(T)P(H) + P(T)P(T)P(T)P(H)
+ P(T)P(T)P(T)P(T)P(T)P(H)
= (1/2)(1/2) + (1/2)(1/2)(1/2)(1/2)
+ (1/2)(1/2)(1/2)(1/2)(1/2)(1/2)
= (1/4) + (1/4)2 + (1/4)3 + (1/4)4 + -------------
= (1/4)/[1 – (1/4)]
= 1/3.

[Ans.]

Q.2. A and B throw two dice simultaneously turn by turn. A will win if he throws a total of 5, B will win if he throws a doublet. Find the probability that B will win the game, though A started it.

Solution :

Let the event of winning A be denoted by ‘A’ and that of B by ‘B’.
P(A) = 4/36 = 1/9 and P(not A) = 1 – 1/9 = 8/9.[(1, 4), (2, 3), (3, 2), (4, 1)]
P(B) = 6/36 = 1/6 and P(notB) = 5/6. [(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)]
Now A starts the game.
Therefore, P(B) = P(not A)P(B) + P(not A)P(not B)P(not A)P(B)
+ P(not A)P(not B)P(not A)P(not B)P(not A)P(B) + --------
= 8/9×1/6 + 8/9×5/6×8/9×1/6 + 8/9×5/6×8/9×5/6×8/9×1/6 + ---------
[This in an infinite G.P.]
a = 8/9×1/6 and r = 5/6×8/9 < 1.
Therefore, P(B) = a/(1 – r)
= [8/9×1/6]/[1 – (5/6×8/9)]
= 8/(54 – 40)
= 8/14 = 4/7.[Ans.]

Q.3. Two dice are rolled once. Find the probability that :

  1. the numbers on two dices are different.
  2. the total of numbers on the two dice is at least 4.

Solution :

Total number of cases are 6×6 = 36.

  1. When the numbers on the two dice are different,
    the number of favourable cases = 6×5 = 30.
    Therefore, probability that the numbers on two dices are different = 30/36 = 5/6. [Ans.]

  2. P(X ≥ 4)
    = 1 – [P(X = 2) + P(X = 3)] [As X = Total no. on two dices & X ≠ 0, 1]
    = 1 – [1/36 + 2/36]
    = 1 – 3/36
    = 33/36 = 11/12.[Ans.]

Maths Paper (With Solutions) By : Mr. M. P. Keshari
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Paper By Mr. M.P.Keshari
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