# Probability Aptitude Questions with Solutions:

#### 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. If the chance of A to pass a exam is $$\frac{2}{3}$$ and chance of B to pass the exam is $$\frac{2}{5}$$, then find the chance that eather A or B pass the exam?

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

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

Solution: Given values, $$P \ (A) = \frac{2}{3}$$

$$P \ (B) = \frac{2}{5}$$

then eather A or B pass the exam, $$P = P \ (A) + P \ (B)$$ $$= \frac{2}{3} + \frac{2}{5}$$ $$= \frac{10 + 6}{15} = \frac{16}{15}$$

1. If three coins are tossed. What is the probability of getting atleast one tail?

1. $$\frac{7}{8}$$
2. $$\frac{8}{7}$$
3. $$\frac{3}{5}$$
4. $$\frac{5}{3}$$

Answer: (a) $$\frac{7}{8}$$

Solution: sample space = $$\{HHH, \ HHT, \ HTT, \ HTH\}$$, $$\{TTT, \ TTH, \ THH, \ THT\}$$

total number of possible outcomes = $$8$$

number of favourable outcomes = $$7$$, then $$Probability \ (P) = \frac{7}{8}$$

1. A fair dice is thrown. What is the probability of getting the number less than $$3 \ ?$$

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

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

Solution: Total numbers on dice = $$\{1, \ 2, \ 3, \ 4, \ 5, \ 6\}$$

numbers on dice less than $$3$$ = $$\{1, \ 2\}$$

so, total number of possible outcomes = $$6$$

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

1. two coins are tossed. What is the probability of getting one head and one tail?

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

Answer: (c) $$\frac{1}{2}$$

Solution: sample space = $$\{HH, \ HT, \ TH, \ TT\}$$

total number of possible outcomes = $$4$$

number of favourable outcomes = $$2$$, then $$Probability \ (P) = \frac{2}{4}$$ $$P = \frac{1}{2}$$

1. A bag contains $$4$$ Black and $$9$$ Red balls. Two balls are drawn from the bag in succession without replacement, then what is the probability that both balls are Red?

1. $$\frac{4}{11}$$
2. $$\frac{3}{9}$$
3. $$\frac{6}{11}$$
4. $$\frac{6}{13}$$

Answer: (d) $$\frac{6}{13}$$

Solution: number of Black balls = $$4$$

number of Red balls = $$9$$, then $$P = \frac{^9C_2}{^{13}C_2}....(1)$$ $$^9C_2 = \frac{9 \times 8}{2} = 36$$ $$^{13}C_2 = \frac{13 \times 12}{2} = 78$$ by putting these values in the equation $$(1)$$, $$P = \frac{36}{78} = \frac{6}{13}$$

1. A bag contains $$3$$ Red and $$5$$ Yellow balls. If two balls from the bag are picked at random. What is the chance that the balls are of different colour?

1. $$\frac{28}{15}$$
2. $$\frac{15}{28}$$
3. $$\frac{28}{13}$$
4. $$\frac{13}{28}$$

Answer: (b) $$\frac{15}{28}$$

Solution: nmber of Red balls = $$3$$

number of Yellow balls = $$5$$, then $$P = \frac{^3C_1 \times ^5C_1}{^8C_2}....(1)$$ where as, $$^3C_1 = 3$$ $$^5C_1 = 5$$ $$^8C_2 = \frac{8 \times 7}{2} = 28$$ by putting the values in equation $$(1)$$, $$P = \frac{3 \times 5}{28} = \frac{15}{28}$$

1. From a pack of $$52$$ playing cards, two cards are drawn at random. What is the chance of drawing one Ace and one King?

1. $$\frac{663}{8}$$
2. $$\frac{8}{668}$$
3. $$\frac{8}{663}$$
4. $$\frac{665}{7}$$

Answer: (c) $$\frac{8}{663}$$

Solution: Total number of cards = $$52$$

number of cards of Ace = $$4$$

number of cards of King = $$4$$, then $$P = \frac{^4C_1 \times ^4C_1}{^{52}C_2}.....(1)$$ where as, $$^4C_1 = 4$$ $$^{52}C_2 = \frac{52 \times 51}{2} = 1326$$ by putting these values in equation $$(1)$$, we get $$P = \frac{4 \times 4}{1326}$$ $$P = \frac{16}{1326} = \frac{8}{663}$$

1. A bag contains $$5$$ Red, $$7$$ Green, and $$9$$ blue balls. If two balls are picked at random, what is the probability of getting that both balls are Green?

1. $$\frac{1}{10}$$
2. $$\frac{10}{1}$$
3. $$\frac{11}{10}$$
4. $$\frac{10}{11}$$

Answer: (a) $$\frac{1}{10}$$

Solution: number of Red balls = $$5$$

number of Green balls = $$7$$

number of Blue balls = $$9$$, then $$P = \frac{^7C_2}{^{21}C_2}.....(1)$$ where as, $$^7C_2 = \frac{7 \times 6}{2} = 21$$ $$^{21}C_2 = \frac{21 \times 20}{2} = 210$$ by putting these values in equation $$(1)$$, we get $$P = \frac{21}{210} = \frac{1}{10}$$

1. If two dice are thrown, then find the probability of getting a total of $$7 \ ?$$

1. $$\frac{2}{3}$$
2. $$\frac{3}{2}$$
3. $$\frac{6}{1}$$
4. $$\frac{1}{6}$$

Answer: (d) $$\frac{1}{6}$$

Solution: Total number of possible outcomes = $$6 \times 6$$ = $$36$$

favourable outcomes = $$(1, \ 6), \ (2, \ 5), \ (3, \ 4)$$, $$(4, \ 3), \ (5, \ 2), (6, \ 1)$$

number of favourable outcomes = $$6$$

then probability of getting a total of $$7$$, $$P = \frac{6}{36} = \frac{1}{6}$$

1. If two dice are tossed. What is the probability that the total is a prime number?

1. $$\frac{5}{12}$$
2. $$\frac{12}{5}$$
3. $$\frac{4}{11}$$
4. $$\frac{11}{4}$$

Answer: (a) $$\frac{5}{12}$$

Solution: there are $$15$$ ways to get a prime number from tossing two dice and prime numbers may be $$2, \ 3, \ 5, \ 7$$ and $$11$$.

total number of possible outcomes = $$6 \times 6$$ = $$36$$

number of favourable outcomes = $$15$$, then $$P = \frac{15}{36} = \frac{5}{12}$$