# Probability Aptitude Questions and Answers:

#### Overview:

 Questions and Answers Type: MCQ (Multiple Choice Questions). Main Topic: Quantitative Aptitude. Quantitative Aptitude Sub-topic: Probability Aptitude Questions and Answers. Number of Questions: 10 Questions with Solutions.

1. A card picked at random from a packed a $$52$$ playing cards. Find the probability that it is a Ace or King?

1. $$\frac{3}{12}$$
2. $$\frac{12}{3}$$
3. $$\frac{2}{13}$$
4. $$\frac{13}{2}$$

Answer: (c) $$\frac{2}{13}$$

Solution: total number of possible outcomes = $$52$$

number of favourable outcomes = $$8$$, then $$Probability = \frac{8}{52}$$ $$= \frac{2}{13}$$

1. A bag contains $$8$$ white and $$9$$ black balls, if two balls are drawn at random what is the probability that both are white?

1. $$\frac{7}{34}$$
2. $$\frac{5}{34}$$
3. $$\frac{5}{27}$$
4. $$\frac{7}{27}$$

Answer: (a) $$\frac{7}{34}$$

Solution: number of white balls = $$8$$

number of black balls = $$9$$, then $$Probability = \frac{^8C_2}{^{17}C_2}$$ $$= \frac{28}{136} = \frac{7}{34}$$

1. If four coins are tossed, then find the probability of getting atleast one tail?

1. $$\frac{16}{15}$$
2. $$\frac{15}{16}$$
3. $$\frac{12}{13}$$
4. $$\frac{13}{12}$$

Answer: (b) $$\frac{15}{16}$$

Solution: total number of events = $$2^4$$ = $$16$$

probability that no tail occurs = $$\frac{1}{16}$$

probability that atleast one tail occurs, $$P = 1 - \frac{1}{16} = \frac{15}{16}$$

1. If a fair dice is thrown, then find the probability of getting a number greater than four?

1. $$\frac{1}{2}$$
2. $$\frac{1}{3}$$
3. $$\frac{1}{4}$$
4. $$\frac{1}{5}$$

Answer: (b) $$\frac{1}{3}$$

Solution: number of possible outcomes = $$6$$

number of favourable events = $$\{5, \ 6\}$$

number of favourable outcomes = $$2$$, then $$Probability = \frac{2}{6}$$ $$= \frac{1}{3}$$

1. If a number is drawn from the numbers between $$1$$ to $$20$$, then find the probability of getting a number multiple of $$4$$?

1. $$\frac{1}{2}$$
2. $$\frac{1}{3}$$
3. $$\frac{1}{4}$$
4. $$\frac{1}{5}$$

Answer: (c) $$\frac{1}{4}$$

Solution: number of possible outcomes = $$20$$

number of favourable events = $$\{4, \ 8, \ 12, \ 16, \ 20\}$$

number of favourable outcomes = $$5$$, then $$Probability = \frac{5}{20}$$ $$= \frac{1}{4}$$

1. If a number is drawn from the numbers between $$1$$ to $$10$$, then find the probability of getting a number multiple of $$3$$?

1. $$\frac{3}{10}$$
2. $$\frac{2}{9}$$
3. $$\frac{3}{13}$$
4. $$\frac{4}{13}$$

Answer: (a) $$\frac{3}{10}$$

Solution: number of possible outcomes = $$10$$

number of favourable events = $$\{3, \ 6, \ 9\}$$

number of favourable outcomes = $$3$$, then $$Probability = \frac{3}{10}$$

1. A card is drawn from the $$52$$ playing cards. Find the probability of getting a heart?

1. $$\frac{1}{2}$$
2. $$\frac{1}{3}$$
3. $$\frac{1}{4}$$
4. $$\frac{1}{5}$$

Answer: (c) $$\frac{1}{4}$$

Solution: number of favourable outcomes = $$13$$

number of possible outcomes = $$52$$, then $$Probability = \frac{13}{52}$$ $$= \frac{1}{4}$$

1. A card is drawn from the $$52$$ playing cards. Find the probability of black colour king?

1. $$\frac{1}{16}$$
2. $$\frac{1}{21}$$
3. $$\frac{1}{13}$$
4. $$\frac{1}{26}$$

Answer: (d) $$\frac{1}{26}$$

Solution: number of favourable outcomes = $$2$$

number of possible outcomes = $$52$$, then $$Probability = \frac{2}{52}$$ $$= \frac{1}{26}$$

1. If probability of A to win a game is $$\frac{5}{7}$$ and probability of B to win the game is $$\frac{3}{5}$$, then find the probability that eather A or B win the game?

1. $$\frac{45}{46}$$
2. $$\frac{46}{45}$$
3. $$\frac{46}{35}$$
4. $$\frac{35}{46}$$

Answer: (c) $$\frac{46}{35}$$

Solution: probability of A to win the game P(A) = $$\frac{5}{7}$$

probability of B to win the game P(B) = $$\frac{3}{5}$$, then $$P = P \ (A) + P \ (B)$$ $$P = \frac{5}{7} + \frac{3}{5}$$ $$= \frac{46}{35}$$

1. If probability of x to pass an exam is $$\frac{1}{5}$$ and probability of y to pass the exam is $$\frac{1}{3}$$, then find the probability that nobody will pass the exam?

1. $$\frac{14}{15}$$
2. $$\frac{15}{14}$$
3. $$\frac{12}{13}$$
4. $$\frac{13}{12}$$

Answer: (a) $$\frac{14}{15}$$

Solution: probability of x to pass the exam P(x) = $$\frac{1}{5}$$

probability of y to pass the exam P(y) = $$\frac{1}{3}$$

then probability that both of them will pass the exam $$P = P \ (x) \times P \ (y)$$ $$P = \frac{1}{5} \times \frac{1}{3}$$ $$= \frac{1}{15}$$ now the probability that nobody will pass the exam, $$= 1 - \frac{1}{15} = \frac{14}{15}$$