CBSE Guess > Papers > Important Questions > Class XII > 2010 > Maths > Mathematics By Mr. M.P.Keshari

Probability

Q.4. Two unbiased dice are tossed simultaneously. Find the probability that the sum of the numbers will be a multiple of 3 or 5.

Solution :

We have, Let event A = Multiple of 3 = {(1, 2), (1, 5), (2, 1), (2, 4), (3, 3), (3, 6),
(4, 2), (4, 5), (5, 1), (5, 4), (6, 3), (6, 6)}
& event B = Multiple of 5 = {(1, 4), (2, 3), (3, 2), (4, 1),
(4, 6), (5, 5)(6, 4)}.
Therefore, P(AUB) = P(A) + P(B) – P(A∩B)
= 12/36 + 7/36 – 0
= 19/36.[Ans.]

Q.5. Two unbiased dice are thrown. Find the probability that the sum of the numbers obtained on the two dice is neither a multiple of 3 nor a multiple of 4.

Solution :

Total number of cases = 6 × 6 = 36.
Let A = event that sum of the numbers obtained on the two dices be a multiple of 3. B = event that sum of the numbers obtained on the two dices be a multiple of 4. Then
A = {(1, 2), (1, 5), (2, 1), (2, 4), (3, 3), (3, 6), (4, 2), (4, 5), (5, 1), (5, 4), (6, 3), (6, 6)}=> o(A) = 12.
B = {(1, 3), (2, 2), (2, 6), (3, 1), (3, 5), (4, 4), (5, 3), (6, 2), (6, 6)}=> o(B) = 9.
Also, A∩B = {(6, 6)}.
Therefore, P(AUB) = P(A) + P(B) – P(A∩B)
= 12/36 + 9/36 – 1/36 = 20/36 = 5/9.
Now, P(Not A∩not B) = P{Not(AUB)}
= 1 – P(AUB)
= 1 – 5/9
= 4/9. [Ans.]

13.2. Multiplication Theorem on Probability/Independent Events.

Q.1. There are two bags. The first bag contains 4 white and 2 black balls, while the second bag contains 3 white and 4 black balls. A ball is picked up at random and a ball is drawn out. Find the probability that it is a white ball.

Solution :

Probability of picking up any bag = 1/2.
(i) If Bag I is picked up, P(White Ball) = 1/2×4/6 = 1/3.
(ii) If Bag II is picked up, P(White Ball) = 1/2×3/7 = 3/14.
The two events ae independent, thus P(White Ball) = 1/3 + 3/7
= 23/42.[Ans.]

Q.2. In a group of 9 students, there are 5 boys and 4 girls. A team of 4 students is to be selected for a quiz competition. Find the probability that there will be 2 boys and 2 girls in that team.

Solution :

We have, no. of boys = 5, no. of girls = 4.
2 boys out of 5 will be selected by 5C2 ways.
2 girls out of 4 will be selected by 4C2 ways.
4 students (2 boys & 2 girls) will be selected out of 9 by 9C4 ways.
Therefore, required probability = (5C2×4C2)/9C4
= (10×6×24)/(9×8×7×6)
= 10/21.[Ans.]

Paper By Mr. M.P.Keshari
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