Chapter 16 Probability Ex 16.3 Solutions
Question - 11 : - In a lottery, a person choses six different natural numbers at random from 1 to 20, and if these six numbers match with the six numbers already fixed by the lottery committee, he wins the prize. What is the probability of winning the prize in the game? [Hint order of the numbers is not important.]
Answer - 11 : -
From the question it is given that,
Total numbers of numbers in the draw = 20
Numbers to be selected = 6
Question - 12 : - Check whether the following probabilities P(A) and P(B) are consistently defined
(i) P(A) = 0.5, P(B) = 0.7, P(A ∩ B) = 0.6
(ii) P(A) = 0.5, P(B) = 0.4, P(A ∪ B) = 0.8
Answer - 12 : -
(i) P(A) = 0.5, P(B) = 0.7, P(A ∩ B) = 0.6
P(A ∩ B) > P(A)
Therefore, the given probabilities are not consistentlydefined.
(ii) P(A) = 0.5, P(B) = 0.4, P(A ∪ B) = 0.8
Then,
P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
0.8 = 0.5 + 0.4 – P(A ∩ B)
Transposing – P(A ∩ B) to LHS and it becomes P(A ∩ B) and0.8 to RHS and it becomes – 0.8.
P(A ∩ B) = 0.9 – 0.8
= 0.1
Therefore, P(A ∩ B) < P(A) and P(A ∩ B) < P(B)
So, the given probabilities are consistently defined.
Question - 13 : - Fill in the blanks in followingtable:
| P(A) | P(B) | P(A ∩ B) | P(A ∪ B) |
| (i) | 1/3 | 1/5 | 1/15 | …. |
| (ii) | 0.35 | ….. | 0.25 | 0.6 |
| (iii) | 0.5 | 0.35 | …. | 0.7 |
Answer - 13 : -
From the given table,
(i) P(A) = 1/3, P(B) = 1/5, P(A ∩ B) = 1/15, P(A ∪ B) = ?
We know that,
P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
= (1/3) + (1/5) – (1/15)
= ((5 + 3)/15) – (1/15)
= (8/15) – (1/15)
= (8 – 1)/15
= 7/15
(ii) P(A) = 0.35, P(B) = ?, P(A ∩ B) = 0.25, P(A ∪ B) = 0.6
Then,
P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
0.6 = 0.35 + P(B) – 0.25
Transposing – 0.25, 0.35 to LHS and it becomes 0.25and –0.35.
P(B) = 0.6 + 0.25 – 0.35
= 0.5
(iii) P(A) = 0.5, P(B) = 0.35, P(A ∪ B) = 0.7, P(A ∩ B) = ?
Then,
P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
0.7 = 0.5 + 0.35 – P(A ∩ B)
Transposing – P(A ∩ B) to LHS and it becomes P(A ∩ B) and0.7 to RHS and it becomes – 0.7.
P(A ∩ B) = 0.85 – 0.7
= 0.15
Question - 14 : - Given P(A) = 5/3 and P(B) = 1/5 . Find P(A or B), if A and B are mutually exclusive events.
Answer - 14 : -
From the question it is given that,
P(A) = 5/3 and P(B) = 1/5
Then, P(A or B), if A and B are mutually exclusive
P(A∪B) or P(A or B) = P(A) + P(B)
= (3/5) + (1/5)
= 4/5
Question - 15 : - If E and F are events such that P(E) = ¼ , P(F) = ½ and P(E and F) = 1/8 , find
(i) P(E or F), (ii) P(not E and not F)
Answer - 15 : -
From the question, we have P(E) = ¼ , P(F) = ½ and P(E ∩ F)= 1/8
(i) P(E or F) i.e. P(E∪F)= P(E) + P(F) – P(E ∩ F)
= ¼ + ½ – (1/8)
= 5/8
= 1 – (5/8)
= (8 – 5)/8
= 3/8
Question - 16 : - Events E and F are such that P(not E or not F) = 0.25, State whether E and F are mutually exclusive.
Answer - 16 : -
Question - 17 : - A and B are events such that P(A) = 0.42, P(B) = 0.48 and P(A and B) = 0.16. Determine (i) P(not A), (ii) P(not B) and (iii) P(A or B)
Answer - 17 : -
From the question it is given that, P(A) = 0.42, P(B) = 0.48and P(A and B) = 0.16.
(i) P(not A) = 1 – P(A)
= 1 – 0.42
= 0.58
(ii) P(not B) = 1 – P(B)
= 1 – 0.48
= 0.52
(iii) P(A not B) = P(A ∪B) = P(A) + P(B) – P(A ∩ B)
= 0.42 + 0.48 – 0.16
= 0.74
Question - 18 : - In Class XI of a school 40% of the students study Mathematics and 30% study Biology. 10% of the class study both Mathematics and Biology. If a student is selected at random from the class, find the probability that he will be studying Mathematics or Biology.
Answer - 18 : -
Question - 19 : - In an entrance test that is graded on the basis of two examinations, the probability of a randomly chosen student passing the first examination is 0.8 and the probability of passing the second examination is 0.7. The probability of passing at least one of them is 0.95. What is the probability of passing both?
Answer - 19 : -
Let us assume probability of a randomly chosen studentpassing the first examination is 0.8 be P(A).
And also assume the probability of passing the secondexamination is 0.7 be P(B)
Then,
P(A∪B) is probability of passing at least one of theexamination
Now,
P(A∪B) = 0.95 , P(A)=0.8, P(B)=0.7
∴ P(A∪B) = P(A) + P(B) – P(A∩B)
0.95 = 0.8 + 0.7 – P(A∩B)
Transposing – P(A ∩ B) to LHS and it becomes P(A ∩ B) and0.95 to RHS and it becomes
– 0.95
P(A∩B) = 1.5 – 0.95
= 0.55
Hence, 0.55 is the probability that student will pass boththe examinations.
Question - 20 : - The probability that a student will pass the final examination in both English and Hindi is 0.5 and the probability of passing neither is 0.1. If the probability of passing the English examination is 0.75, what is the probability of passing the Hindi examination?
Answer - 20 : -
Let us assume probability of passing the English examinationis 0.75 be P(A).
And also assume the probability of passing the Hindiexamination is be P(B).
Here given, P(A) = 0.75, P(A∩B) – 0.5, P(AI∩BI)= 0.1
We know that, P(AI∩BI) = 1 – P(A∪B)
Then, P(A∪B) = 1 – P(AI∩BI)
= 1 – 0.1
= 0.9
∴ P(A∪B) = P(A) + P(B) – P(A∩B)
0.9 = 0.75 + P(B) – 0.5
Transposing 0.75, – 0.5 to LHS and it becomes – 0.75, 0.5.
P(B) = 0.9 + 0.5 – 0.75
= 0.65