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