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. |

- If two coins are tossed, then find the probability of obtaining two heads?
- \(\frac{1}{2}\)
- \(\frac{2}{3}\)
- \(\frac{1}{3}\)
- \(\frac{1}{4}\)

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

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

total number of possible outcomes = \(4\)

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

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

total number of possible outcomes = \(4\)

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

- If a fair dice is thrown, then find the probability of a number greater than three?
- \(\frac{1}{2}\)
- \(\frac{1}{3}\)
- \(\frac{1}{4}\)
- \(\frac{1}{5}\)

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 $$ Probability \ (P) = \frac{3}{6} $$ $$ = \frac{1}{2} $$

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

total number of possible outcomes = \(6\)

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

- Two coins are tossed. Find the probability of obtaining one head and one tail?
- \(\frac{1}{4}\)
- \(\frac{1}{3}\)
- \(\frac{1}{2}\)
- \(\frac{2}{3}\)

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

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

total number of possible outcomes = \(4\)

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

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

total number of possible outcomes = \(4\)

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

- If a fair dice is thrown, then find the probability of obtaining number \(5 \ ?\)
- \(\frac{1}{3}\)
- \(\frac{1}{6}\)
- \(\frac{1}{2}\)
- \(\frac{1}{5}\)

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 $$ Probability \ (P) = \frac{1}{6} $$

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

total number of possible outcomes = \(6\)

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

- 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?
- \(\frac{1}{6}\)
- \(\frac{2}{5}\)
- \(\frac{2}{7}\)
- \(\frac{3}{5}\)

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 \((1)\), $$ Probability \ (P) = \frac{6}{21}$$ $$ = \frac{2}{7} $$

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 \((1)\), $$ Probability \ (P) = \frac{6}{21}$$ $$ = \frac{2}{7} $$

- 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?
- \(\frac{1}{2}\)
- \(\frac{1}{20}\)
- \(\frac{1}{24}\)
- \(\frac{1}{22}\)

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 \((1)\), $$ Probability \ (P) = \frac{20}{440} $$ $$ = \frac{1}{22} $$

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 \((1)\), $$ Probability \ (P) = \frac{20}{440} $$ $$ = \frac{1}{22} $$

- A fair dice is thrown. Find the probability of obtaining a odd number?
- \(\frac{1}{2}\)
- \(\frac{1}{3}\)
- \(\frac{1}{4}\)
- \(\frac{3}{5}\)

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 $$ Probability \ (P) = \frac{3}{6} $$ $$ = \frac{1}{2} $$

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 $$ Probability \ (P) = \frac{3}{6} $$ $$ = \frac{1}{2} $$

- If two dice are thrown. Find the probability of obtaining the total of \(12 \ ?\)
- \(\frac{1}{22}\)
- \(\frac{1}{32}\)
- \(\frac{1}{36}\)
- \(\frac{1}{30}\)

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 $$ Probability \ (P) = \frac{1}{36} $$

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 $$ Probability \ (P) = \frac{1}{36} $$

- If two dice are thrown, then find the probability of getting a total of \(3 \ ?\)
- \(\frac{1}{24}\)
- \(\frac{1}{20}\)
- \(\frac{1}{12}\)
- \(\frac{1}{18}\)

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 $$ Probability \ (P) = \frac{2}{36} $$ $$ = \frac{1}{18} $$

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 $$ Probability \ (P) = \frac{2}{36} $$ $$ = \frac{1}{18} $$

- 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?
- \(\frac{3}{25}\)
- \(\frac{2}{25}\)
- \(\frac{3}{33}\)
- \(\frac{2}{27}\)

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, $$ Probability \ (P) = \frac{2}{5} \times \frac{3}{10} $$ $$ = \frac{3}{25} $$

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, $$ Probability \ (P) = \frac{2}{5} \times \frac{3}{10} $$ $$ = \frac{3}{25} $$

Lec 1: Introduction to Probability
Exercise-1
Lec 2: Addition Rule
Exercise-2
Lec 3: Multiplication Rule
Exercise-3
Exercise-4
Exercise-5