Practicing the AP Board Solutions Class 11 Maths and Chapter 14 Inter 1st Year Maths Probability Solutions Exercise 14b Pdf Download will help students to clear their doubts quickly.
Intermediate 1st Year Maths Probability Solutions Exercise 14b
Probability Exercise 14b Solutions
Probability Class 11 Exercise 14b Solutions – Probability 14b Exercise Solutions
I.
Question 1.
A coin is tossed twice. What is the probability that atleast one tail occurs?
Solution:
When a coin is tossed twice
S = {HH, HT, TH, TT}
n(s) = 4
“Atleast one tail” means 1 tailor, 2 tails
E = {HT, TH, TT}
n(E) = 3
P(E) = \(\frac{\mathrm{n}(\mathrm{E})}{\mathrm{n}(\mathrm{~S})}=\frac{3}{4}\)
Question 2.
If \(\frac {2}{11}\) is the probability of an event A, then what is the probability of the event ‘not A’?
Solution:
P(A) = \(\frac {2}{11}\)
P(not A) = 1 – P(A)
= 1 – \(\frac {2}{11}\)
= \(\frac {9}{11}\)
Question 3.
A letter is chosen at random from the letters of the word ‘ASSASSINATION’. Find the probability that this letter is (i) a vowel, (ii) a consonant.
Solution:
ASSASSINATION
Total letters = 13
Vowels: A, I, O
A: 3 times, I: 2 times, O: 1 time
Total vowels = 3 + 2 + 1 = 6
Consonants: S, S, S, S, N, N, T
Count = 13 – 6 = 7
(i) Probability of a vowel
P(E) = \(\frac{\mathrm{n}(\mathrm{E})}{\mathrm{n}(\mathrm{~S})}=\frac{6}{13}\)
(ii) Probability of a consonant
P(E) = \(\frac{\mathrm{n}(\mathrm{E})}{\mathrm{n}(\mathrm{~S})}=\frac{7}{13}\)
Question 4.
Given P(A) = \(\frac {3}{5}\) and P(B) = \(\frac {1}{5}\). Find P(A or B) if A and B are mutually exclusive events.
Solution:
Given P(A) = \(\frac {3}{5}\), P(B) = \(\frac {1}{5}\)
If A and B are mutually exclusive, then
P(A or B) = P(A ∪ B)
= P(A) + P(B)
= \(\frac {3}{5}\) + \(\frac {1}{5}\)
= \(\frac {4}{5}\)
II.
Question 1.
Which of the following can not be a valid assignment of probabilities for outcomes of the sample space S = {ω1, ω2, ω3, ω4, ω5, ω6, ω7}
Solution:
Here, each of the numbers P(ω1) is positive and less than 1.
Sum of probabilities = P(ω1) + P(ω2) + P(ω3) + P(ω4) + P(ω5) + P(ω6) + P(ω7)
= 0.1 + 0.01 + 0.05 + 0.03 + 0.01 + 0.2 + 0.6
= 1
Thus, the assignment is valid.
Question 2.
A die is thrown. Find the probability of the following events:
(i) A prime number will appear
(ii) A number greater than or equal to 3 will appear
(iii) A number less than or equal to one will appear
(iv) A number more than 6 will appear
(v) A number less than 6 will appear
Solution:
S = {1, 2, 3, 4, 5, 6}
Total out comes = n(S) = 6
(i) Prime numbers on die = {2, 3, 5}
n(E) = 3
P(E) = \(\frac{\mathrm{n}(\mathrm{E})}{\mathrm{n}(\mathrm{~S})}=\frac{3}{6}=\frac{1}{2}\)
(ii) A number greater than or equal to 3 will appear = {3, 4, 5, 6}
n(E) = 4
P(E) = \(\frac{\mathrm{n}(\mathrm{E})}{\mathrm{n}(\mathrm{~S})}=\frac{4}{6}=\frac{2}{3}\)
(iii) A number less than or equal to will appear {1}.
n(E) = 1
P(E) = \(\frac {1}{6}\)
(iv) A number more than 6 will appear as ‘0.’
n(E) = 0
P(E) = \(\frac {0}{6}\) = 0
(v) A number less than 6 will appear {1, 2, 3, 4, 5}
n(E) = 5
P(E) = \(\frac{\mathrm{n}(\mathrm{E})}{\mathrm{n}(\mathrm{~S})}=\frac{5}{6}\)
Question 3.
A card is selected at random from a pack of 52 cards.
(a) How many points are there in the sample space?
(b) Calculate the probability that the card is an ace of spades.
(c) Calculate the probability that the card is (i) an ace, (ii) black.
Solution:
(a) When a card is selected from a pack of 52 cards, the number of possible outcomes is 52
i.e., the sample space contains 52 elements.
Therefore, there are 52 points in the sample space.
(b) Let A be the event in which the card drawn is an ace of spades.
Accordingly, n(A) = 1
∴ P(A) = \(\frac{\text { Number of outcomes favourable to A }}{\text { Total number of possible outcomes }}\)
= \(\frac{\mathrm{n}(\mathrm{~A})}{\mathrm{n}(\mathrm{~S})}\)
= \(\frac {1}{52}\)
(c) (i) Let E be the event in which the card drawn is an ace.
Since there are 4 aces in a pack of 52 cards, n(E) = 4.
∴ P(E) = \(\frac{\text { Number of outcomes favourable to E }}{\text { Total number of possible outcomes }}\)
= \(\frac{\mathrm{n}(\mathrm{~E})}{\mathrm{n}(\mathrm{~S})}\)
= \(\frac {4}{52}\)
= \(\frac {1}{13}\)
(ii) Let F be the event in which the card drawn is black.
Since there are 26 black cards in a pack of 52 cards, n(F) = 26.
∴ P(F) = \(\frac{\text { Number of outcomes favourable to F }}{\text { Total number of possible outcomes }}\)
= \(\frac{\mathrm{n}(\mathrm{~F})}{\mathrm{n}(\mathrm{~S})}\)
= \(\frac {26}{52}\)
= \(\frac {1}{2}\)
Question 4.
A fair coin with 1 marked on one face and 6 on the other, and a fair die are both tossed. Find the probability that the sum of numbers that turn up is (i) 3, (ii) 12.
Solution:
Coin outcomes: {1, 6} → 2 outcomes
Die outcomes: {1, 2, 3, 4, 5, 6} → 6 outcomes
Total Number of outcomes = 2 × 6 = 12
S = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
(i) Probability that the sum is 3
coin + die = 3
Only (1, 2) is possible
P(sum = 3) = \(\frac {1}{2}\)
(ii) Probability that the sum is 12
coin + die = 12
Only (6, 6) is possible
P(sum = 12) = \(\frac {1}{2}\)
Question 5.
There are four men and six women on the city council. If one council member is selected for a committee at random, how likely is it that it is a woman?
Solution:
There are four men and six women on the city council.
As one council member is to be selected for a committee at random, the sample space contains (4 + 6) elements.
Let A be the event in which the selected council member is a woman.
Accordingly, n(A) = 6
∴ P(B) = \(\frac{\text { Number of outcomes favourable to B }}{\text { Total number of possible outcomes }}\)
= \(\frac{\mathrm{n}(\mathrm{~B})}{\mathrm{n}(\mathrm{~S})}\)
= \(\frac {6}{10}\)
= \(\frac {3}{5}\)
Question 6.
In a lottery, a person chooses six different natural numbers at random from 1 to 20, and if these six numbers match 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]
Solution:
These are 20 natural numbers from 1 to 20
Total number = 20
Number to be choosen = 6
Number of ways choosing 6 natural numbers from 1 to 20 = 20C6
= \(\frac{20!}{6!(20-6)!}\)
= \(\frac{20!}{6!4!}\)
= 38760 ways
Question 7.
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
Solution:
(i) Given P(A) = 0.5, P(B) = 0.7, P(A ∩ B) = 0.6
We have P(A ∩ B) ≤ min(P(A), P(B))
Here P(A ∩ B) = 0.6 > P(A) = 0.5
This is not possible; the intersection cannot be greater than the individual event in a consistent manner.
(ii) P(A) = 0.5, P(B) = 0.4, P(A ∩ B) = 0.8
P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
⇒ 0.8 = 0.5 + 0.4 – P(A ∩ B)
⇒ P(A ∩ B) = 0.9 – 0.8 = 0.1
0 ≤ P (A ∩ B) ≤ min (P(A), P(B)) = 0.4
∴ Consistent.
Question 8.
Fill in the blanks in the following table:
Solution:
(i) Here, P(A) = \(\frac {1}{3}\), P(B) = \(\frac {1}{5}\), P(A ∩ B) = \(\frac {1}{5}\)
P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
= \(\frac{1}{3}+\frac{1}{5}-\frac{1}{15}\)
= \(\frac{5+3-1}{15}\)
= \(\frac {7}{15}\)
(ii) Here, P(A) = 0.35, P(A ∩ B) = 0.25, P(A ∪ B) = 0.6
We know that P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
⇒ 0.6 = 0.35 + P(B) – 0.25
⇒ P(B) = 0.6 – 0.35 + 0.25
⇒ P(B) = 0.5
(iii) Here, P(A) = 0.5, P(B) = 0.35, P(A ∪ B) = 0.7
We know that P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
⇒ 0.7 = 0.5 + 0.35 – P(A ∩ B)
⇒ P(A ∩ B) = 0.5 + 0.35 – 0.7
⇒ P(A ∩ B) = 0.15
Question 9.
If E and F are events such that P(E) = \(\frac {1}{4}\), P(F) = \(\frac {1}{2}\) and P(E and F) = \(\frac {1}{8}\), find (i) P(E or F), (ii) P(not E and not F).
Solution:
Here, P(E) = \(\frac {1}{4}\), P(F) = \(\frac {1}{2}\), and P(E and F) = \(\frac {1}{8}\)
(i) We know that P(E or F) = P(E) + P(F) – P(E and F)
∴ P(E or F) = \(\frac{1}{4}+\frac{1}{2}-\frac{1}{8}=\frac{2+4-1}{8}=\frac{5}{8}\)
(ii) We have E’ ∩ F’ = (E ∪ F)’
By De Morgan’s law
∴ P(E’ ∪ F’) = P(E ∪ F)’
= 1 – P(E ∪ F)
= 1 – \(\frac {5}{8}\)
= \(\frac {3}{8}\)
∴ P(E’ ∩ F’) = \(\frac {3}{8}\)
Thus, P(not E and not F) = \(\frac {3}{8}\)
Question 10.
Events E and F are such that P(not E or not F) = 0.25. State whether E and F are mutually exclusive.
Solution:
P(not E and not F) = 0.25
P(Ec ∩ Fc) = 0.25
P(Ec ∩ Fc) = 1 – P(E ∪ F)
= 1 – 0.25
= 0.75
∴ P(E ∪ F) = P(E) + P(F) – P(E ∩ F)
0.75 = P(E) + P(F) – P(E ∩ F)
If E and F were mutually exclusive, then P(E ∩ F) = 0.
then P(E ∪ F) = P(E) + P(F)
P(E) + P(F) = 0.75
We cannot directly determine whether P(E ∩ F) = 0 or not because we do not know P(E), P(F), and P(E ∩ F) independently.
Question 11.
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).
Solution:
It is given that P(A) = 0.42, P(B) = 0.48, 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) We know that P(A or B) = P(A) + P(B) – P(A and B)
= 0.42 + 0.48 – 0.16
= 0.74
III.
Question 1.
Three coins are tossed once. Find the probability of getting
(i) 3 heads
(ii) 2 heads
(iii) atleast 2 heads
(iv) at most 2 heads
(v) no head
(vi) 3 tails
(vii) exactly two tails
(viii) no tail
(ix) at most two tails
Solution:
When three coins are tossed once, the sample space is given by
S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}
⇒ n(S) = 8
It is known that the probability of an event A is given by
∴ P(A) = \(\frac{\text { Number of outcomes favourable to A }}{\text { Total number of possible outcomes }}\)
= \(\frac{\mathrm{n}(\mathrm{~A})}{\mathrm{n}(\mathrm{~S})}\)
(i) Let B be the event of the occurrence of 3 heads.
Accordingly B = {HHH}
∴ P(B) = \(\frac{\mathrm{n}(\mathrm{~B})}{\mathrm{n}(\mathrm{~S})}=\frac{1}{8}\)
(ii) Let C be the event of the occurrence of 2 heads.
Accordingly, C = {HHT, HTH, THH}
∴ P(C) = \(\frac{\mathrm{n}(\mathrm{C})}{\mathrm{n}(\mathrm{~S})}=\frac{3}{8}\)
(iii) Let D be the event of the occurrence of at least 2 heads.
Accordingly, D = {HHH, HHT, HTH, THH}
∴ P(D) = \(\frac{\mathrm{n}(\mathrm{D})}{\mathrm{n}(\mathrm{~S})}=\frac{4}{8}=\frac{1}{2}\)
(iv) Let E be the event of the occurrence of at most 2 heads.
Accordingly, E = {HHT, HTH, THH, HTT, THT, TTH, TTT}
∴ P(E) = \(\frac{\mathrm{n}(\mathrm{E})}{\mathrm{n}(\mathrm{~S})}=\frac{7}{8}\)
(v) Let F be the event of the occurrence of no heads.
Accordingly F = {TTT}
∴ P(F) = \(\frac{\mathrm{n}(\mathrm{~F})}{\mathrm{n}(\mathrm{~S})}=\frac{1}{8}\)
(vi) Let G be the event of the occurrence of 3 tails.
Accordingly G = {TTT}
∴ P(G) = \(\frac{n(G)}{n(S)}=\frac{1}{8}\)
(vii) Let H be the event of the occurrence of exactly 2 tails.
Accordingly, H = {HTT, THT, TTH}
∴ P(H) = \(\frac{\mathrm{n}(\mathrm{H})}{\mathrm{n}(\mathrm{~S})}=\frac{3}{8}\)
(viii) Let I be the event of the occurrence of no tail.
Accordingly, I = {HHH}
∴ P(I) = \(\frac{\mathrm{n}(\mathrm{H})}{\mathrm{n}(\mathrm{~S})}=\frac{1}{8}\)
(ix) Let J be the event of the occurrence of at most 2 tails.
Accordingly, J = {HHH, HHT, HTH, THH, HTT, THT, TTH}
∴ P(J) = \(\frac{\mathrm{n}(\mathrm{~J})}{\mathrm{n}(\mathrm{~S})}=\frac{7}{8}\)
Question 2.
In Class XI of a school 40% of the students study Mathematics and 30% study Biology. Ten % of the class studies 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.
Solution:
P(Mathematics) = 40% = 0.40
P(Biology) = 30% = 0.30
P(Both Mathematics and Biology) = 10% = 0.10
∴ P(M ∪ B) = P(M) + P(B) – P(M ∩ B)
= 0.40 + 0.30 – 0.10
= 0.60 (or) 60%
Question 3.
In an entrance test that is graded based on 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 atleast one of them is 0.95. What is the probability of passing both?
Solution:
Probability of passing the first examination
P(A) = 0.8
Probability of passing the second examination
P(B) = 0.7
Probability of passing atleast one exam
P(A ∪ B) = 0.95
P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
⇒ 0.95 = 0.5 + 0.7 – P(A ∩ B)
⇒ 0.95 = 1.5 – P(A ∩ B)
⇒ P(A ∩ B) = 1.5 – 0.95
⇒ P(A ∩ B) = 0.55 or 55%
Question 4.
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?
Solution:
P(E ∩ H) = 0.5
P(neither) = 0.1
P(E) = 0.75
P(E ∪ H) = 1 – P(neither)
= 1 – 0.1
= 0.9
P(E ∪ H) = P(E) + P(H) – P(E ∩ H)
⇒ 0.9 = 0.75 + P(H) – 0.5
⇒ 0.9 = 0.25 + P(H)
⇒ P(H) = 0.9 – 0.25
⇒ P(H) = 0.65 (or) 65%
Question 5.
In a class of 60 students, 30 opted for NCC, 32 opted for NSS, and 24 opted for both NCC and NSS. If one of these students is selected at random, find the probability that
(i) The student opted for NCC or NSS.
(ii) The student has opted for neither NCC nor NSS.
(iii) The student has opted for NSS but not NCC.
Solution:
Let A be the event in which the selected student has opted for NCC, and B be the event in which the selected student has opted for NSS.
Total number of students = 60
Number of students who have opted for NCC = 30
∴ P(A) = \(\frac{30}{60}=\frac{1}{2}\)
Number of students who have opted for NSS = 32
∴ P(B) = \(\frac{32}{60}=\frac{8}{15}\)
Number of students who have opted for both NCC and NSS = 24
∴ P(A and B) = \(\frac{24}{60}=\frac{2}{5}\)
(i) We know that P(A or B) = P(A) + P(B) – P(A and B)
∴ P(A and B) = \(\frac{1}{2}+\frac{8}{15}-\frac{2}{5}\)
= \(\frac{15+16-12}{30}\)
= \(\frac {19}{30}\)
(ii) P(not A and not B) = P(A’ and B’)
= P(A’ ∩ B’)
= P[(A ∪ B)’]
= 1 – P(A ∪ B)
= 1 – P(A or B)
= 1 – \(\frac {19}{30}\)
= \(\frac {11}{30}\)
Thus, the probability that the selected student has opted neither NCC nor NSS is \(\frac {11}{30}\).
(iii) The given information can be represented by a Venn diagram as,
It is clear that, number of students who have opted for NSS but not NCC = n(B – A)
= n(B) – n(A ∩ B)
= 32 – 24
= 8
Thus, the probability that the selected student has opted for NSS but not for NCC = \(\frac{8}{60}=\frac{2}{15}\)
Question 6.
A fair coin is tossed four times, and a person wins Rs. 1.00 for each head and loses Rs. 1.50 for each tail that turns up. From the sample space, calculate how many different amounts of money you can have after four tosses and the probability of having each of these amounts.
Solution:
A fair coin is tossed 4 times
Head (H) = win Rs. 1.00
Tail (T) = lose Rs. 1.50
Total number of outcomes in 4 tosses
24 = 16
h = number of heads
t = 4 – h = number of tails
Net amount = (1.00 × h) + (-1.50 × t) = h – 1.5(4 – h)
Net amount = h – 6 + 1.5h = 2.5h – 6
Net amount for each possible number of heads (0 to 4)
