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 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?
- \(\frac{15}{16}\)
- \(\frac{16}{15}\)
- \(\frac{12}{13}\)
- \(\frac{13}{12}\)

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, $$ P = P \ (A) + P \ (B) $$ $$ = \frac{2}{3} + \frac{2}{5} $$ $$ = \frac{10 + 6}{15} = \frac{16}{15} $$

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

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

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

- If three coins are tossed. What is the probability of getting atleast one tail?
- \(\frac{7}{8}\)
- \(\frac{8}{7}\)
- \(\frac{3}{5}\)
- \(\frac{5}{3}\)

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 $$ Probability \ (P) = \frac{7}{8} $$

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

total number of possible outcomes = \(8\)

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

- A fair dice is thrown. What is the probability of getting the number less than \(3 \ ?\)
- \(\frac{1}{2}\)
- \(\frac{1}{3}\)
- \(\frac{1}{5}\)
- \(\frac{1}{4}\)

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

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

- two coins are tossed. What is the probability of getting one head and one tail?
- \(\frac{1}{4}\)
- \(\frac{1}{3}\)
- \(\frac{1}{2}\)
- \(\frac{1}{5}\)

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

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

total number of possible outcomes = \(4\)

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

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

total number of possible outcomes = \(4\)

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

- 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?
- \(\frac{4}{11}\)
- \(\frac{3}{9}\)
- \(\frac{6}{11}\)
- \(\frac{6}{13}\)

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 \((1)\), $$ P = \frac{36}{78} = \frac{6}{13} $$

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 \((1)\), $$ P = \frac{36}{78} = \frac{6}{13} $$

- 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?
- \(\frac{28}{15}\)
- \(\frac{15}{28}\)
- \(\frac{28}{13}\)
- \(\frac{13}{28}\)

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 \((1)\), $$ P = \frac{3 \times 5}{28} = \frac{15}{28} $$

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 \((1)\), $$ P = \frac{3 \times 5}{28} = \frac{15}{28} $$

- From a pack of \(52\) playing cards, two cards are drawn at random. What is the chance of drawing one Ace and one King?
- \(\frac{663}{8}\)
- \(\frac{8}{668}\)
- \(\frac{8}{663}\)
- \(\frac{665}{7}\)

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 $$ P = \frac{4 \times 4}{1326} $$ $$ P = \frac{16}{1326} = \frac{8}{663} $$

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 $$ P = \frac{4 \times 4}{1326} $$ $$ P = \frac{16}{1326} = \frac{8}{663} $$

- 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?
- \(\frac{1}{10}\)
- \(\frac{10}{1}\)
- \(\frac{11}{10}\)
- \(\frac{10}{11}\)

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 $$ P = \frac{21}{210} = \frac{1}{10} $$

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 $$ P = \frac{21}{210} = \frac{1}{10} $$

- If two dice are thrown, then find the probability of getting a total of \(7 \ ?\)
- \(\frac{2}{3}\)
- \(\frac{3}{2}\)
- \(\frac{6}{1}\)
- \(\frac{1}{6}\)

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 \(7\), $$ P = \frac{6}{36} = \frac{1}{6} $$

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 \(7\), $$ P = \frac{6}{36} = \frac{1}{6} $$

- If two dice are tossed. What is the probability that the total is a prime number?
- \(\frac{5}{12}\)
- \(\frac{12}{5}\)
- \(\frac{4}{11}\)
- \(\frac{11}{4}\)

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 $$ P = \frac{15}{36} = \frac{5}{12} $$

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 $$ P = \frac{15}{36} = \frac{5}{12} $$

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