Multiplication theorem on probability, 20 mcqs on it with detailed explained answers
Here are 20 MCQs on Multiplication Theorem on Probability with detailed explained answers:
1. If $$P(A) = \frac{1}{3}$$ and $$P(B|A) = \frac{1}{4}$$, find $$P(A \cap B)$$.
- a) $$\frac{1}{7}$$
- b) $$\frac{1}{12}$$
- c) $$\frac{1}{9}$$
- d) $$\frac{1}{6}$$
Answer: $$P(A \cap B) = P(A) \times P(B|A) = \frac{1}{3} \times \frac{1}{4} = \frac{1}{12}$$.
2. Two cards are drawn without replacement from a pack. Probability both are aces?
- a) $$\frac{1}{221}$$
- b) $$\frac{1}{169}$$
- c) $$\frac{1}{52}$$
- d) $$\frac{1}{435}$$
Answer: $$ \frac{4}{52} \times \frac{3}{51} = \frac{12}{2652} = \frac{1}{221}$$.
3. If events A and B are independent and $$P(A) = 0.5$$, $$P(B) = 0.3$$, find $$P(A \cap B)$$.
- a) 0.8
- b) 0.15
- c) 0.1
- d) 0.5
Answer: $$P(A \cap B) = P(A) \times P(B) = 0.5 \times 0.3 = 0.15$$.
4. If $$P(A) = 0.6$$ and $$P(B|A) = 0.2$$, find $$P(A \cap B)$$.
- a) 0.12
- b) 0.3
- c) 0.2
- d) 0.6
Answer: $$P(A \cap B) = 0.6 \times 0.2 = 0.12$$.
5. Probability of drawing a red card and then a king when drawing 2 cards without replacement?
- a) $$\frac{1}{17}$$
- b) $$\frac{1}{26}$$
- c) $$\frac{1}{52}$$
- d) $$\frac{1}{34}$$
Answer: Probability red card = 26/52 = 1/2. After that king: 2 kings left out of 51 cards → $$1/2 \times 2/51 = 1/51$$.
6. If $$P(A) = 0.4$$, $$P(B|A) = 0.5$$, find $$P(A \cap B)$$.
- a) 0.2
- b) 0.5
- c) 0.1
- d) 0.9
Answer: $$0.4 \times 0.5 = 0.2$$.
7. Two dice rolled. Probability first die is 4 and second is 5?
- a) $$\frac{1}{12}$$
- b) $$\frac{1}{36}$$
- c) $$\frac{1}{18}$$
- d) $$\frac{1}{6}$$
Answer: Since independent, multiply single probabilities: $$\frac{1}{6} \times \frac{1}{6} = \frac{1}{36}$$.
8. If events A and B are independent with $$P(A) = 0.3$$, $$P(B) = 0.7$$, find $$P(B|A)$$.
- a) 0.3
- b) 0.7
- c) 0.21
- d) 1
Answer: For independent events, $$P(B|A) = P(B) = 0.7$$.
9. Probability of selecting a white ball and then a black ball from bag with 3 white and 2 black balls without replacement?
- a) $$\frac{3}{10}$$
- b) $$\frac{6}{10}$$
- c) $$\frac{3}{5}$$
- d) $$\frac{2}{5}$$
Answer: $$P = \frac{3}{5} \times \frac{2}{4} = \frac{3}{10}$$.
10. $$P(A) = 0.5$$, $$P(B|A) = 0.3$$. Find probability both A and B do not occur.
- a) 0.35
- b) 0.65
- c) 0.7
- d) 0.85
Answer: $$P(A \cap B) = 0.5 \times 0.3 = 0.15$$, so $$P((A \cap B)') = 1 - 0.15 = 0.85$$.
11. Probability of drawing an ace and then a king with replacement from a deck?
- a) $$\frac{1}{169}$$
- b) $$\frac{1}{78}$$
- c) $$\frac{1}{52}$$
- d) $$\frac{1}{221}$$
Answer: $$P = \frac{4}{52} \times \frac{4}{52} = \frac{16}{2704} = \frac{1}{169}$$.
12. For independent events A and B, if $$P(A \cup B) = 0.74$$, and $$P(A) = 0.4$$, $$P(B) = 0.5$$, find $$P(A \cap B)$$.
- a) 0.1
- b) 0.2
- c) 0.3
- d) 0.4
Answer: $$P(A \cap B) = P(A) + P(B) - P(A \cup B) = 0.4 + 0.5 - 0.74 = 0.16$$.
13. Probability of selecting 3 white balls consecutively from 5 white and 7 black balls without replacement?
- a) $$\frac{5}{12} \times \frac{4}{11} \times \frac{3}{10}$$
- b) $$\frac{5}{12} \times \frac{5}{11} \times \frac{5}{10}$$
- c) $$\frac{5}{12} \times \frac{7}{11} \times \frac{6}{10}$$
- d) $$\frac{5}{12} \times \frac{4}{12} \times \frac{3}{12}$$
Answer: $$= \frac{5}{12} \times \frac{4}{11} \times \frac{3}{10} = \frac{60}{1320} = \frac{1}{22}$$.
14. Probability first drawing a red card and then a heart without replacement?
- a) $$\frac{1}{4}$$
- b) $$\frac{1}{8}$$
- c) $$\frac{13}{52} \times \frac{12}{51}$$
- d) $$\frac{26}{52} \times \frac{13}{51}$$
Answer: $$P = \frac{26}{52} \times \frac{13}{51} = \frac{1}{2} \times \frac{13}{51} = \frac{13}{102}$$.
15. If $$P(A) = 0.8$$, $$P(B|A) = 0.25$$, find $$P(A^c \cap B^c)$$.
- a) 0.15
- b) 0.2
- c) 0.6
- d) 0.4
Answer: $$P(A \cap B) = 0.8 \times 0.25 = 0.2$$, so $$P(A \cup B)^c = 1 - P(A \cup B) = ?$$ But $$P(A^c \cap B^c) = 1 - P(A \cup B) \geq 1 - (0.8 + 0.25) = negative$$, but since max is 1, exact calculation requires more info. Assuming independence for simplicity: $$P(A^c \cap B^c) = (1-0.8)(1-0.25) = 0.2 \times 0.75 = 0.15$$.
16. Probability of getting tail and then head when tossing a coin twice?
- a) 0.25
- b) 0.5
- c) 0.75
- d) 0.1
Answer: $$P = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4} = 0.25$$.
17. Probability that two dice show a sum of 9 if the events are independent?
- a) 1/9
- b) 1/12
- c) 1/8
- d) 1/36
Answer: Sum of 9 has 4 favorable pairs out of 36, so $$P=4/36=1/9$$.
18. If $$P(A) = 0.4$$, $$P(B|A) = 0.3$$, $$P(B|A^c) = 0.2$$, find $$P(B)$$.
- a) 0.24
- b) 0.26
- c) 0.3
- d) 0.32
Answer: By total probability, $$P(B) = P(A) P(B|A) + P(A^c) P(B|A^c) = 0.4 \times 0.3 + 0.6 \times 0.2 = 0.12 + 0.12 = 0.24$$.
19. Probability that none of two cards drawn without replacement are spades?
- a) $$\frac{39}{52} \times \frac{38}{51}$$
- b) $$\frac{13}{52} \times \frac{12}{51}$$
- c) $$\frac{39}{52} \times \frac{13}{51}$$
- d) $$\frac{38}{52} \times \frac{39}{51}$$
Answer: Cards not spade = 39, so $$P = \frac{39}{52} \times \frac{38}{51}$$.
20. Probability first drawing a 3 and then drawing a 4 from a pack with replacement?
- a) $$\frac{1}{676}$$
- b) $$\frac{1}{169}$$
- c) $$\frac{1}{52}$$
- d) $$\frac{1}{26}$$
Answer: $$P = \frac{4}{52} \times \frac{4}{52} = \frac{1}{13} \times \frac{1}{13} = \frac{1}{169}$$.
