Very Short Answer on Probability Class 11 OP Malhotra Exe-22G ISC Maths Solutions Ch-22. In this article you would learn to solve hard problems easily on Probability. Step by step solutions of latest textbook has been given as latest syllabus. Visit official Website CISCE for detail information about ISC Board Class-11.
Probability Class 11 OP Malhotra Very Short Answer ISC Maths Solutions Ch-22
| Board | ICSE |
| Publications | S Chand |
| Subject | Maths |
| Class | 11th |
| Chapter-22 | Probability |
| Writer | OP Malhotra |
| Exe-22(G) | Very Short Answer Type Questions. |
Very Short Answer on Probability
OP Malhotra ISC Class 11 Maths Solutions
Que-1: A coin is tossed twice. The probability of at least one tail occurring is ________
Sol: Sample space = {HH, HT, TH, TT}
Favourable outcomes (at least one tail) = {HT, TH, TT} = 3
Total outcomes = 4
Probability = 3/4
Que-2: A card is selected from a pack of 52 cards. The probability of getting a black card is ________
Sol: Total cards = 52
Black cards (spades + clubs) = 26
Probability = 26/52 = 1/2
Que-3: Three coins are tossed once. The probability of getting at most two tails = ________
Sol: Total outcomes = 8
“At most two tails” means 0, 1, or 2 tails
Only case excluded = TTT
Favourable outcomes = 7
Probability = 7/8
Que-4: Given P(A) = 3/5 and P(B) = 1/5, find P(A or B) if A and B are mutually exclusive events.
Sol: For mutually exclusive events:
P(A ∪ B) = P(A) + P(B)
= 3/5 + 1/5 = 4/5
Que-5: In throwing two fair dice, find the probability of obtaining a sum greater than 3 but not exceeding 6.
Sol: Possible sums: 4, 5, 6
Ways for 4 = 3, for 5 = 4, for 6 = 5
→ total = 12
Total outcomes = 36
Probability = 12/36 = 1/3
Que-6: A bag contains 17 tickets numbered 1 to 17. Two tickets are drawn without replacement. Find the probability that both are even numbers.
Sol: Even numbers = 8
P(first even) = 8/17
P(second even) = 7/16
Probability = (8/17) × (7/16)
= 56/272 = 7/34
Que-7: Find the probability of obtaining a sum 8 in a single throw of two dice.
Sol: Possible pairs: (2,6), (3,5), (4,4), (5,3), (6,2) = 5
Total outcomes = 36
Probability = 5/36
Que-8: A bag contains 4 red, 6 white, and 5 black balls. Two balls are drawn. Find the probability of getting one red and one white ball.
Sol: Total balls = 15
Ways = (4×6) × 2 = 48
Total ways = 15×14 = 210
Probability = 48/210 = 8/35
Que-9: If P(A) = 0.5 and P(B) = 0.3 and A, B are mutually exclusive, find probability of neither A nor B.
Sol: P(A ∪ B) = 0.5 + 0.3 = 0.8
P(neither) = 1 − 0.8 = 0.2
Que-10: If e₁, e₂, e₃, e₄ are outcomes and P(e₁)=0.1, P(e₂)=0.5, P(e₃)=0.1, find P(e₄).
Sol: Total probability = 1
P(e₄) = 1 − (0.1 + 0.5 + 0.1) = 0.3
Que-11: A card is drawn from a well-shuffled standard deck of cards. What is the probability that the card is either an ace or a jack or a king.
Sol: Aces = 4, Jacks = 4, Kings = 4
→ total = 12
Probability = 12/52 = 3/13
Que-12: A box contains 3 red balls and 3 green balls. Two balls are picked up consecutively without replacement. If a green ball is picked up in the first turn, what is the probability of picking up a green ball again in the second turn?
Sol: Total balls = 6
Ball picked up = 1
Remaining balls = 5, green left = 2
Probability = 2/5
= 0.4
Que-13: If A and B are two events associated with an experiment such that P(A ∪ B) = P(A ∩ B) and P(A) = 1/3, find P(B).
Sol: We know the formula:
P(A ∪ B) = P(A) + P(B) − P(A ∩ B)
Given: P(A ∪ B) = P(A ∩ B)
So,
P(A) + P(B) − P(A ∩ B) = P(A ∩ B)
⇒ P(A) + P(B) = 2P(A ∩ B)
But since P(A ∪ B) = P(A ∩ B), this is possible only when:
A = B (both events are identical)
Hence,
P(B) = P(A) = 1/3
Que-14: A bag has 5 red balls and some blue balls. If probability of drawing blue is double that of red balls, then find number of blue balls.
Sol: Let blue = x
P(red) = 5/(5+x),
P(blue) = x/(5+x)
Given x/(5+x) = 2×[5/(5+x)]
⇒ x = 10
Que-15: Twelve tickets are numbered from 1 to 12. One ticket is drawn at rando, then find probability of number to be divisible by 2 or 3.
Sol: Multiples of 2 = 6, of 3 = 4, common = 2
Total favourable = 6 + 4 − 2 = 8
Probability = 8/12
= 2/3
–: End Probability Class 11 OP Malhotra Exe-22G ISC Maths Ch-22 Solutions :–
Return to :- OP Malhotra ISC Class-11 S Chand Publication Maths Solutions
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