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. |

- A card picked at random from a packed a \(52\) playing cards. Find the probability that it is a Ace or King?
- \(\frac{3}{12}\)
- \(\frac{12}{3}\)
- \(\frac{2}{13}\)
- \(\frac{13}{2}\)

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

Solution: total number of possible outcomes = \(52\)

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

Solution: total number of possible outcomes = \(52\)

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

- A bag contains \(8\) white and \(9\) black balls, if two balls are drawn at random what is the probability that both are white?
- \(\frac{7}{34}\)
- \(\frac{5}{34}\)
- \(\frac{5}{27}\)
- \(\frac{7}{27}\)

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

Solution: number of white balls = \(8\)

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

Solution: number of white balls = \(8\)

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

- If four coins are tossed, then find the probability of getting atleast one tail?
- \(\frac{16}{15}\)
- \(\frac{15}{16}\)
- \(\frac{12}{13}\)
- \(\frac{13}{12}\)

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, $$ P = 1 - \frac{1}{16} = \frac{15}{16} $$

Solution: total number of events = \(2^4\) = \(16\)

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

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

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

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

Solution: number of possible outcomes = \(6\)

number of favourable events = \(\{5, \ 6\}\)

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

Solution: number of possible outcomes = \(6\)

number of favourable events = \(\{5, \ 6\}\)

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

- If a number is drawn from the numbers between \(1\) to \(20\), then find the probability of getting a number multiple of \(4\)?
- \(\frac{1}{2}\)
- \(\frac{1}{3}\)
- \(\frac{1}{4}\)
- \(\frac{1}{5}\)

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 $$ Probability = \frac{5}{20} $$ $$ = \frac{1}{4} $$

Solution: number of possible outcomes = \(20\)

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

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

- If a number is drawn from the numbers between \(1\) to \(10\), then find the probability of getting a number multiple of \(3\)?
- \(\frac{3}{10}\)
- \(\frac{2}{9}\)
- \(\frac{3}{13}\)
- \(\frac{4}{13}\)

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

Solution: number of possible outcomes = \(10\)

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

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

Solution: number of possible outcomes = \(10\)

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

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

- A card is drawn from the \(52\) playing cards. Find the probability of getting a heart?
- \(\frac{1}{2}\)
- \(\frac{1}{3}\)
- \(\frac{1}{4}\)
- \(\frac{1}{5}\)

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

Solution: number of favourable outcomes = \(13\)

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

Solution: number of favourable outcomes = \(13\)

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

- A card is drawn from the \(52\) playing cards. Find the probability of black colour king?
- \(\frac{1}{16}\)
- \(\frac{1}{21}\)
- \(\frac{1}{13}\)
- \(\frac{1}{26}\)

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

Solution: number of favourable outcomes = \(2\)

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

Solution: number of favourable outcomes = \(2\)

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

- 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?
- \(\frac{45}{46}\)
- \(\frac{46}{45}\)
- \(\frac{46}{35}\)
- \(\frac{35}{46}\)

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 $$ P = P \ (A) + P \ (B) $$ $$ P = \frac{5}{7} + \frac{3}{5} $$ $$ = \frac{46}{35} $$

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 $$ P = P \ (A) + P \ (B) $$ $$ P = \frac{5}{7} + \frac{3}{5} $$ $$ = \frac{46}{35} $$

- 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?
- \(\frac{14}{15}\)
- \(\frac{15}{14}\)
- \(\frac{12}{13}\)
- \(\frac{13}{12}\)

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, $$ = 1 - \frac{1}{15} = \frac{14}{15} $$

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, $$ = 1 - \frac{1}{15} = \frac{14}{15} $$

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