Practice AP Inter 2nd Year Maths Study Material Chapter 13 Probability MCQ to identify your strengths and weak areas.
AP Inter 2nd Year Maths Probability MCQ
Question 1.
If P(A) = \(\frac{1}{2}\), P(B) = (), then P(A|B) is
1) 0
2) \(\frac{1}{2}\)
3) not exist
4) 1
Solution:
3) not exist
P(A/B) = \(\frac{\mathrm{P}(\mathrm{~A} \cap \mathrm{~B})}{\mathrm{P}(\mathrm{~B})}=\frac{\mathrm{P}(\mathrm{~A} \cap \mathrm{~B})}{0}\) which is not defined
Question 2.
If A and B are events such that P(A|B) = P(B|A), then
1) A ⊂ B but A ≠ B
2) A = B
3) A ∩ B = Φ
4) P(A) = P(B)
Solution:
Given P(A/B) = P(B/A)
⇒ \(\frac{\mathrm{P}(\mathrm{~A} \cap \mathrm{~B})}{\mathrm{P}(\mathrm{~B})}=\frac{\mathrm{P}(\mathrm{~B} \cap \mathrm{~A})}{\mathrm{P}(\mathrm{~A})} \Rightarrow \frac{\mathrm{P}(\mathrm{~A} \cap \mathrm{~B})}{\mathrm{P}(\mathrm{~B})}=\frac{\mathrm{P}(\mathrm{~A} \cap \mathrm{~B})}{\mathrm{P}(\mathrm{~A})}\) [∵ A ∩ B = B ∩ A]
⇒ \(\frac{1}{P(B)}=\frac{1}{P(A)}\) ⇒ P(A) = P(B)
Question 3.
The probability of obtaining an even prime number on each die, when a pair of dice is rolled is
1) 0
2) \(\frac{1}{3}\)
3) \(\frac{1}{12}\)
4) \(\frac{1}{36}\)
Solution:
4) \(\frac{1}{36}\)
When two dice are rolled, the number of outcomes n(S) = 62 = 36.
The only even prime number is 2.
Let E be the event of getting an even prime number on each die. ∴ E = (2, 2) ⇒ P(E) = \(\frac{1}{36}\)
Question 4.
Two events A and B will be independent, if
1) A and B are mutually exclusive
2) P(A’B’) = [1 – P(A)] [1 – P(B)]
3) P(A) = P(B)
4) P(A) + P(B) = 1
Solution:
2) P(A’B’) = [1 – P(A)] [1 – P(B)]
A and B are independent ⇒ A’ and B’ are independent
⇒ P(A’ ∩ B’) = [1 – P(A)] [1 – P(B)] are independent
Question 5.
If A and B are two events such that A ⊂ B and P(B) ≠ 0, then which of the following is correct?
1) P(A|B) = \(\frac{\mathrm{P}(\mathrm{~B})}{\mathrm{P}(\mathrm{~A})}\)
2) P(A|B) < P(A)
3) P(A|B) ≥ P(A)
4) P(A) = P(B)
Solution:
3) P(A|B) ≥ P(A)
If A ⊂ B, then A ∩ B ⇒ P(A ∩ B) = P(A). Also, P(A) < P(B)
P(A|B) = \(\frac{\mathrm{P}(\mathrm{~A} \cap \mathrm{~B})}{\mathrm{P}(\mathrm{~B})}=\frac{\mathrm{P}(\mathrm{~A})}{\mathrm{P}(\mathrm{~B})}\) …..(1)
Since P(B) ≤ 1 ⇒ \(\frac{1}{\mathrm{P}(\mathrm{~B})} \geq 1 \Rightarrow \frac{\mathrm{P}(\mathrm{~A})}{\mathrm{P}(\mathrm{~B})} \geq \mathrm{P}(\mathrm{~A})\)
From (1), we have P(A|B) ≥ P(A)
Question 6.
If A and B are two events such that P(A) ≠ 0 and P(B | A) = I, then
1) A ⊂ B
2) B ⊂ A
3) B = Φ
4) A = Φ
Solution:
1) A ⊂ B
Given P(A) ≠ 0 and P(B|A) = 1,
∴ \(P(B \mid A)=\frac{P(B \cap A)}{P(A)} \Rightarrow 1=\frac{P(B \cap A)}{P(A)}\) ⇒ P(A) = P(B ∩ A) ⇒ A = A ∩ B ⇒ A⊂ B
Question 7.
If P(A|B) > P(A), then which of the following is correct :
1) P(B|A) < P(B)
2) P(A ∩ B) < P(A) . P(B) 3) P(B|A) > P(B)
4) P(B|A) = P(B)
Solution:
3) P(B|A) > P(B)
Given that
Given, P(A|B) > P(A) ⇒ \(\frac{\mathrm{P}(\mathrm{~A} \cap \mathrm{~B})}{\mathrm{P}(\mathrm{~B})}>\mathrm{P}(\mathrm{~A})\)
⇒ P(A ∩ B) > P(A) × P(B) ⇒ \(\frac{\mathrm{P}(\mathrm{~A} \cap \mathrm{~B})}{\mathrm{P}(\mathrm{~A})}\) ⇒ P(B) ⇒ P(B|A) > P(B)
Question 8.
If A and B are any two events such that P(A) + P(B) – P(A and B) = P(A), then
1) P(B|A) = 1
2) P(A|B) = 1
3) P(B|A) = 0
4)P(A|B) = 0
Solution:
2) P(A|B) = 1
Given that P(A) + P(B) – P(A and B) =P(A),
⇒ P(A) + P(B) – P(A ∩ B) = P(A) ⇒ P(B) – P(A ∩ B) = 0 ⇒ P(A ∩ B) = P(B)
P(A|B) = \(\frac{\mathrm{P}(\mathrm{~A} \cap \mathrm{~B})}{\mathrm{P}(\mathrm{~B})}=\frac{\mathrm{P}(\mathrm{~B})}{\mathrm{P}(\mathrm{~B})}\) = 1
Question 9.
A bag contains 7 red and 3 white balls. Three balls are drawn one after other without replacement. Then the probability that the first two are red and third one is white is
1) \(\frac{7}{40}\)
2) \(\frac{33}{40}\)
3) \(\frac{23}{40}\)
4) \(\frac{17}{40}\)
Solution:
1) \(\frac{7}{40}\)
7R + 3W = Total 10 balls
P(E) = P(Red and Red and White) = \(\left(\frac{7}{10}\right) \times \frac{6}{9} \times \frac{3}{8}=\frac{7}{40}\) (∵ Drawn ball is not replaced)
Question 10.
A book consists «f 20 pages. If two pages arc drawn (opened) at random, then the probability that both numbers are prime numbers is
1) \(\frac{17}{95}\)
2) \(\frac{16}{95}\)
3) \(\frac{2}{15}\)
4) \(\frac{14}{95}\)
Solution:
4) \(\frac{14}{95}\)
Total no. of pages = 20
Primes up to 20 are 2, 3, 5, 7, 11, 13, 17, 19 & the no. of these primes = 8
∴ P(E) = \(\frac{{ }^8 \mathrm{C}_2}{{ }^{20} \mathrm{C}_2}=\frac{8 \times 7}{20 \times 19}=\frac{2 \times 7}{5 \times 19}=\frac{14}{95}\)
Question 11.
A fair coin is tossed 3 times, then the probability of getting one head and two tails is
1) \(\frac{1}{8}\)
2) \(\frac{1}{4}\)
3) \(\frac{3}{8}\)
4) \(\frac{1}{2}\)
Solution:
3) \(\frac{3}{8}\)
S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}
E = {HTT, THT, TTH} ⇒ n(E) = 3; n(S) = 8 P(E) = \(\frac{\mathrm{n}(\mathrm{E})}{\mathrm{n}(\mathrm{~S})}=\frac{3}{8}\)
Question 12.
A person appears lor an interview for two posts A and B. The selection of the posts are independent. If P( A) = \(\frac{1}{5}\), P(B) = \(\frac{1}{8}\) then P(A ∪ B) is
1) \(\frac{7}{10}\)
2) \(\frac{3}{10}\)
3) \(\frac{9}{10}\)
4) \(\frac{1}{10}\)
Solution:
2) \(\frac{3}{10}\)
Given A, B are independent event ⇒ P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
= \(\frac{1}{5}+\frac{1}{8}\) – [P(A).P(B)] = \(\frac{13}{40}-\left(\frac{1}{5} \times \frac{1}{8}\right)=\frac{13}{40}-\frac{1}{40}=\frac{12}{40}=\frac{3}{10}\)
Question 13.
Three events A, B, C are mutually exclussive and exhaustive and P(A) = 0.4, then P(B) + P(C) =
1) 0.4
2) 0.5
3) 0.6
4) 0
Solution:
3) 0.6
A, B, C are mutually exclusive and exhaustive ⇒ A ∪ B ∪ C = S ……..(1)
Given P(A) = 0.4 ……..(2)
(1) ⇒ P(A ∪ B ∪ C) = P(S) ⇒ P(A) + P(B) + P(C) = 1 ⇒ 0.4 + P(B) + P(C) = 1
⇒ P(B) + P(C) = 1 – 0.4 = 0.6
Question 14.
If P(A ∪ B) = 0.65 and P(A ∩ B) = 0.15, then P(\(\vec{A}\)) + P(\(\vec{B}\)) =3 J
1) 0.8
2) 0.6
3) 1.2
4) 1.4
Solution:
3) 1.2
\(\mathrm{P}(\overline{\mathrm{~A}})+\mathrm{P}(\overline{\mathrm{~B}})\) = 1 – P(A) + 1 – P(B)
= 2 – [P(A) + P(B)] = 2 – [P(A ∪ B) + P(A ∩ B)]
= 2 – [0.65 + 0.15] = 2 – [0.80] = 1.20 = 1.2
Question 15.
When two dice are rolled, the probability of getting unequal numbers on the faces is
1) \(\frac{1}{6}\)
2) \(\frac{35}{36}\)
3) \(\frac{5}{6}\)
4) \(\frac{1}{3}\)
Solution:
3) \(\frac{5}{6}\)
Two dice are rolled n(S) = 36
Equal number faces = {(1, 1) (2, 2) (3, 3) (4, 4)(5, 5)(6, 6)} ⇒ n(S) = 6
⇒ no.of unequal faces = 36 – 6 = 30 = n(E)
∴ P(E) = \(\frac{\mathrm{n}(\mathrm{E})}{\mathrm{n}(\mathrm{~S})}=\frac{30}{36}=\frac{5}{6}\)
Question 16.
A fair coin whose faces are marked with I and 2 is thrown for four times. Then the probability of throwing a total of atleast 5 is
1) \(\frac{1}{16}\)
2) \(\frac{5}{16}\)
3) \(\frac{15}{16}\)
4) \(\frac{3}{16}\)
Solution:
3) \(\frac{15}{16}\)
Coin with faces 1. (say H); 2. (say T)
thrown 4 – times
Getting total at least 5 ⇒ total ≥ 5 ⇒ x ≥ 5
Now, P(x ≥ 5) = 1 – P(x < 5) = 1 – P (getting a total 4 faces 4 – times)
= 1 – P(every time a face 1) = 1 – \(\left[\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2}\right]=1-\frac{1}{16}=\frac{16-1}{16}=\frac{15}{16}\)
Question 17.
If A and B are independent events of an experiment then which of the following statements is true
1) P(A ∪ B) = P(A) + P(B) – P(A) . P(B)
2) P(A|B) = P(A) and P(B|A) = P(B)
3) P(A ∩ B) = P(A) . P(B)
4) All the above
Solution:
4) All the above
By definition P(A ∩ B) = P(A).P(B)
Question 18.
If A and B are two events of a random experiment of throwing a die given by “A” : throwing an odd face and B : throwing a composite face.
Then which of the following statements is correct. ?
1)A and Bare equally likely
2) A and B are mutually exclusive
3) A and B are mutually exhaustive
4) A and B are linearly independent
Solution:
2) A and B are mutually exclusive
When a die is thrown
A : odd face (1, 3, 5); B : composite face (4, 6). Then P(A) =\(\frac{3}{6}=\frac{1}{2}\) and P(B) = \(\frac{2}{6}=\frac{1}{3}\)
1) A, B are likely (✗)
2) A, B are mutually exclusive(✓)
A ∩ B = Φ (or) P(A ∩ B) = 0
3) A, B mutually exhaustive (✗) ∵ A ∪ B ≠ S
4) P(A ∩ B) ≠ P(A).P(B) ⇒ NOT independent (✗)
Question 19.
If A, B and C are independent events of a random experiment such that P(A) = p, P(B) = q, P(C) = r, where p, a, r ∈ (0, 1). Then the probability of the event A only occurs is
1) p . q . r
2) p(1 – q)(1 – r)
3) p. q(1 – r)
4) (1 – p) (1 – q) (1 – r)
Solution:
2) p(1 – q)(1 – r)
A, B, C are independent
P(A only occurs) = \(\mathrm{P}(\mathrm{~A} \cap \overline{\mathrm{~B}} \cap \overline{\mathrm{C}})=\mathrm{P}(\mathrm{~A}) \cdot \mathrm{P}(\overline{\mathrm{~B}}) \cdot \mathrm{P}(\overline{\mathrm{C}})\) (∵ A, B, C are independent)
= P(A).[1 – P(B)][1 – P(C)] = P[1 – q][1 – r]
Question 20.
A fair die is rolled. Consider the events A = {I, 3, 5} and B = {2, 3}, then P(A|B) is
1) \(\frac{1}{2}\)
2) \(\frac{1}{3}\)
3) \(\frac{2}{3}\)
4) \(\frac{5}{6}\)
Solution:
1) \(\frac{1}{2}\)
S = {1, 2, 3, 4, 8, 6}; A = {1, 3, 5}, B = {2, 3} ⇒ P(B) =\(\frac{2}{6}=\frac{1}{3}\)
∴ (A ∩ B) = {3} ⇒ P(A ∩ B) = \(\frac{1}{6}\)
∴ P(A|B) = \(\frac{\mathrm{P}(\mathrm{~A} \cap \mathrm{~B})}{\mathrm{P}(\mathrm{~B})}=\frac{1 / 6}{1 / 3}=\frac{1}{6} \times \frac{3}{1}=\frac{1}{2}\)