# 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. If two coins are tossed, then find the probability of obtaining two heads?

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

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

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

total number of possible outcomes = $$4$$

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

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

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

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

Solution: numbers on dice greater than three = $$\{4, \ 5, \ 6\}$$

total number of possible outcomes = $$6$$

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

1. Two coins are tossed. Find the probability of obtaining one head and one tail?

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

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

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

total number of possible outcomes = $$4$$

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

1. If a fair dice is thrown, then find the probability of obtaining number $$5 \ ?$$

1. $$\frac{1}{3}$$
2. $$\frac{1}{6}$$
3. $$\frac{1}{2}$$
4. $$\frac{1}{5}$$

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

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

total number of possible outcomes = $$6$$

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

1. A bag contains $$3$$ black and $$4$$ white balls. If two balls picked at random, then what is the probability that both balls are white?

1. $$\frac{1}{6}$$
2. $$\frac{2}{5}$$
3. $$\frac{2}{7}$$
4. $$\frac{3}{5}$$

Answer: (c) $$\frac{2}{7}$$

Solution: number of black balls = $$3$$

number of white balls = $$4$$, then $$P = \frac{^4C_2}{^7C_2}....(1)$$ $$^4C_2 = \frac{4 \times 3}{2} = \frac{12}{2} = 6$$ $$^7C_2 = \frac{7 \times 6}{2} = \frac{42}{2} = 21$$ by putting the values of $$^4C_2$$ and $$^7C_2$$ in equestion $$(1)$$, $$Probability \ (P) = \frac{6}{21}$$ $$= \frac{2}{7}$$

1. Three balls are to be drawn from a bag containing $$5$$ Red and $$7$$ white balls. Find the probability that all the three balls will be Red?

1. $$\frac{1}{2}$$
2. $$\frac{1}{20}$$
3. $$\frac{1}{24}$$
4. $$\frac{1}{22}$$

Answer: (d) $$\frac{1}{22}$$

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

number of White balls = $$7$$, then $$P = \frac{^5C_3}{^{12}C_3}.....(1)$$ $$^5C_3 = \frac{5 \times 4 \times 3}{3} = 20$$ $$^{12}C_3 = \frac{12 \times 11 \times 10}{3} = 440$$ by putting the values of $$^5C_3$$ and $$^{12}C_3$$ in equestion $$(1)$$, $$Probability \ (P) = \frac{20}{440}$$ $$= \frac{1}{22}$$

1. A fair dice is thrown. Find the probability of obtaining a odd number?

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

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

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

odd numbers on dice = $$\{1, \ 3, \ 5\}$$

total number of possible outcomes = $$6$$

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

1. If two dice are thrown. Find the probability of obtaining the total of $$12 \ ?$$

1. $$\frac{1}{22}$$
2. $$\frac{1}{32}$$
3. $$\frac{1}{36}$$
4. $$\frac{1}{30}$$

Answer: (c) $$\frac{1}{36}$$

Solution: there is only one way to obtain a total of $$12$$ = $$\{6, \ 6\}$$

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

number of favourable outcome = $$1$$, then $$Probability \ (P) = \frac{1}{36}$$

1. If two dice are thrown, then find the probability of getting a total of $$3 \ ?$$

1. $$\frac{1}{24}$$
2. $$\frac{1}{20}$$
3. $$\frac{1}{12}$$
4. $$\frac{1}{18}$$

Answer: (d) $$\frac{1}{18}$$

Solution: there are only two conditions to obtain the total $$3$$ = $$\{(1, \ 2)(2, \ 1)\}$$

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

number of favourable outcomes = $$2$$, then $$Probability \ (P) = \frac{2}{36}$$ $$= \frac{1}{18}$$

1. Two friends appeared in an test. If probability of the first friend to pass the test is $$\frac{2}{5}$$ and second friend is $$\frac{3}{10}$$, then find what is the probability that both of them will be passed the test?

1. $$\frac{3}{25}$$
2. $$\frac{2}{25}$$
3. $$\frac{3}{33}$$
4. $$\frac{2}{27}$$

Answer: (a) $$\frac{3}{25}$$

Solution: probability of first friend to pass the test = $$\frac{2}{5}$$

probability of second friend to pass the test = $$\frac{3}{10}$$

then probability that both will be passed, $$Probability \ (P) = \frac{2}{5} \times \frac{3}{10}$$ $$= \frac{3}{25}$$