RowQ
The Vault
RowQ
The Vault
CBSE Class 11 Maths · 10 questions · 24 marks
This chapter reframes chance as arithmetic: every random experiment has a sample space of equally likely outcomes, and the probability of an event is simply how large a slice of that space it occupies. Building on your set theory from Chapter 1, you will compute probabilities of unions and complements using the same Venn-diagram logic, applied now to coins, dice, and cards instead of abstract sets. This axiomatic approach is the foundation for all probability and statistics you will study later.
A die is thrown once. The probability of getting a number greater than 4 is:
Answer
The probability is 1/3. Sample space S = {1,2,3,4,5,6}, so n(S) = 6. Numbers greater than 4 are {5,6}, so n(E) = 2. P(E) = 2/6 = 1/3.
Two coins are tossed simultaneously. The probability of getting at least one head is:
Answer
The probability is 3/4. Sample space S = {HH, HT, TH, TT}, so n(S) = 4. At least one head means all outcomes except TT, so n(E) = 3. P(E) = 3/4.
If P(A) = 0.4, P(B) = 0.5 and P(A ∩ B) = 0.2, then P(A ∪ B) equals:
Answer
P(A ∪ B) = 0.7. Using the addition theorem: P(A∪B) = P(A) + P(B) - P(A∩B) = 0.4 + 0.5 - 0.2 = 0.7.
A card is drawn at random from a well-shuffled deck of 52 cards. The probability that it is a king or a queen is:
Answer
The probability is 2/13. There are 4 kings and 4 queens, so 8 favourable cards out of 52 (kings and queens cannot overlap, so they're mutually exclusive). P = 8/52 = 2/13.
Assertion (A): If P(A) = 0.6, then P(not A) = 0.4. Reason (R): For any event A, P(A) + P(A') = 1.
Answer
Both A and R are true and R is the correct explanation of A. From P(A)+P(A')=1 with P(A)=0.6, we get P(A')=1-0.6=0.4, matching Assertion A, which is true. Reason R states exactly the complement rule used to derive this value, so R correctly explains A.
A bag contains 5 red balls and 7 black balls. One ball is drawn at random. Find the probability that it is red.
Answer
Total balls = 5 + 7 = 12. Favourable outcomes (red balls) = 5. P(red) = 5/12.
Two dice are thrown together. Find the probability that the sum of the numbers appearing is 8.
Answer
Sample space has n(S) = 36 equally likely outcomes. Pairs summing to 8: (2,6), (3,5), (4,4), (5,3), (6,2) — that is 5 outcomes. P(sum = 8) = 5/36.
In a class of 60 students, 30 study Physics, 25 study Chemistry, and 12 study both. A student is selected at random. Find the probability that the student studies (i) Physics or Chemistry, (ii) neither subject, (iii) exactly one of the two subjects.
Answer
Step 1 — Let P = event studies Physics, C = event studies Chemistry. P(P) = 30/60 = 1/2, P(C) = 25/60 = 5/12, P(P∩C) = 12/60 = 1/5. Step 2 — (i) P(P∪C) = P(P)+P(C)-P(P∩C) = 1/2 + 5/12 - 1/5. LCD of 2,12,5 is 60: = 30/60 + 25/60 - 12/60 = 43/60. Step 3 — (ii) P(neither) = 1 - P(P∪C) = 1 - 43/60 = 17/60. Step 4 — (iii) Exactly one = P(P∪C) - P(P∩C) = 43/60 - 12/60 = 31/60. So the probabilities are 43/60, 17/60, and 31/60 respectively.
A committee of 4 people is to be chosen at random from a group of 7 men and 5 women. Find the probability that the committee has (i) exactly 2 women, (ii) at least 3 women.
Answer
Step 1 — Total ways to choose 4 from 12 people: 12C4 = (12×11×10×9)/(4×3×2×1) = 11880/24 = 495. Step 2 — (i) Exactly 2 women means 2 women from 5 and 2 men from 7: 5C2 × 7C2 = 10 × 21 = 210. P(exactly 2 women) = 210/495 = 14/33. Step 3 — (ii) At least 3 women means 3 women+1 man, or 4 women+0 men. 3 women, 1 man: 5C3 × 7C1 = 10 × 7 = 70. 4 women, 0 men: 5C4 × 7C0 = 5 × 1 = 5. Total favourable = 70 + 5 = 75. P(at least 3 women) = 75/495 = 5/33. So the probabilities are 14/33 and 5/33 respectively.
Read the following and answer the questions that follow: A survey of 100 office workers found that 55 drink tea, 40 drink coffee, and 20 drink both tea and coffee. One worker is selected at random from this group. (a) Find the probability that the worker drinks tea. (b) Find the probability that the worker drinks tea or coffee. (c) Find the probability that the worker drinks neither tea nor coffee. (d) Find the probability that the worker drinks exactly one of the two drinks.
Answer
(a) P(tea) = 55/100 = 11/20. (b) P(tea∪coffee) = P(tea)+P(coffee)-P(both) = 55/100+40/100-20/100 = 75/100 = 3/4. (c) P(neither) = 1 - 75/100 = 25/100 = 1/4. (d) P(exactly one) = P(tea∪coffee) - P(both) = 75/100 - 20/100 = 55/100 = 11/20.
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