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This is one of our many quizzes on Multiplication rule. Click on this to access more such competitive quizzes on Permutations and Combinations.

Multiplication rule Quiz 2

mathematics

 If in a city X, 40% of the population like mango juice, 25% like banana juice and 15% like both mango and banana juice. If a person is selected at random from the city X, like mango juice, the probability that he also likes banana juice is.

 First find the probability of the population who like mango juice, banana juice and both of them separately, then use the formula$P\left( A/B \right)=\dfrac{P\left( A\cap B \right)}{P\left( B \right)}$.

  $   \text{Let }  MJ= \text{ Mango Juice, } {BJ= }\text{ Banana juice}\text{.} \\  \text{According to the question, } \\   P\left( MJ \right)=40\%\text{ =}0.4\text{ and }P\left( BJ \right)=25\%=0.25 \\   P\left( MJ\text{ and }BJ \right)=15\%=0.15 \\   \text{Hence required probability, } \\   P\left( MJ/BJ \right)=\dfrac{P\left( MJ\cap BJ \right)}{P\left( BJ \right)}=\dfrac{0.15}{0.25} \\   \therefore P\left( MJ/BJ \right)=\dfrac{3}{5}=0.60\text{ }\!\!~\!\!\text{  or }60\% \\ $

 A bag contains 3 Apples and 4 Mangoes. Another bag contains 1 Apple and 4 Mangoes. A bag and a fruit are picked at random. What is the probability that the fruit will be Apple?

 First find the probability of selecting a bag, then an apple from any of the two bags separately, then add them for the total probability.

$ &nbsp;&nbsp;&nbsp;\text{The probability of selecting one bag=}\dfrac{1}{2} \\ &nbsp;&nbsp;\text{Probability of getting an apple from the 1st bag=}\dfrac{3}{7} \\ &nbsp;&nbsp;\text{Probability of getting an apple from 2nd bag=}\dfrac{1}{5} \\ &nbsp;&nbsp;\text{Total probability of getting an apple=}\dfrac{1}{2}\times \dfrac{3}{7}+\dfrac{1}{2}\times \dfrac{1}{5}=\dfrac{11}{35} \\ $

 It is known that a student speaks truth in 60% of his statements. He has given an objective type exam with 10 questions, 1 mark for each correct answer and no negative marks. The teacher allowed him to evaluate his answers and report his mark by himself. What is the probability that he actually scored 10 marks in the exam?

 First find the probability of the event when the student scored 10 marks, when he reports his score is 10 marks separately, then use the formula $P\left( A/B \right)=\dfrac{P\left( A\cap B \right)}{P\left( B \right)}$.

$\begin{array}{*{35}{l}} &nbsp;&nbsp;&nbsp;\text{Let A be the event that the student scored 10 marks}\text{.} \\ &nbsp;&nbsp;&nbsp;\text{And, B is the event that he reports his score as 10 marks}\text{.} \\ &nbsp;&nbsp;&nbsp;\text{His score can be any one out of 0, 1, 2, 3,10 marks}\text{.} \\ &nbsp;&nbsp;&nbsp;\text{i}\text{.e}\text{. Such 11 possible scores}\text{.} \\ &nbsp;&nbsp;&nbsp;\text{Therefore }P\left( A \right)=\dfrac{1}{11} \\ &nbsp;&nbsp;&nbsp;\text{One out of the total 11 cases of event A, the student reported truly } \\ &nbsp;&nbsp;\text{and the remaining 10 cases he reported falsely}\text{.} \\ &nbsp;&nbsp;\\ &nbsp;&nbsp;&nbsp;\text{Therefore, }P\left( B \right)=\dfrac{1}{11}\times \dfrac{3}{5}+\dfrac{10}{11}\times \dfrac{2}{5}=\dfrac{23}{55} \\ &nbsp;&nbsp;&nbsp;\text{And, }P\left( AB \right)=\dfrac{1}{11}\times \dfrac{3}{5}=\dfrac{3}{55} \\ &nbsp;&nbsp;&nbsp;\text{The probability that the student actually scored 10 marks in the exam is;} \\ &nbsp;&nbsp;&nbsp;P\left( A/B \right)=\dfrac{P\left( A\cap B \right)}{P\left( B \right)}=\dfrac{\dfrac{3}{55}}{\dfrac{23}{55}}=\dfrac{3}{23} \\ \end{array}$

 A teacher is hired by a school to teach a class. She already knows that 10% of the children in the class are academically weak, denoted by W, whereas 90% of the class is academically strong, denoted by S. The probability that an academically weak child has poor grades is 0.95. But, occasionally, an academically strong child also has poor grades, with conditional probability 0.08. Upon her interaction with one of the children in the class, she finds that the child has poor grades. What is the probability that the child is academically weak?

 First write all the conditional probabilities given then use Bayes’ theorem, $P\left( W/A \right)=\dfrac{P\left( W \right)P\left( A/W \right)}{P\left( W \right)P\left( A/W \right)+P\left( S \right)P\left( A/S \right)}$

$\begin{array}{*{35}{l}} &nbsp;&nbsp;&nbsp;\text{Let A be the event that the child has poor grades}\text{.} \\ &nbsp;&nbsp;&nbsp;P\left( W \right)=0.9\text{ and }P\left( S \right)=0.1\text{ } \\ &nbsp;&nbsp;&nbsp;P\left( A/W \right)=0.95\text{ and }P\left( A/S \right)=0.08 \\ &nbsp;&nbsp;&nbsp;\text{Thus, using Bayes Theorem,} \\ &nbsp;&nbsp;&nbsp;\dfrac{0.1\times 0.95}{0.1\times 0.95+0.9\times 0.08} \\ &nbsp;&nbsp;&nbsp;=\dfrac{0.095}{0.095+0.072} \\ &nbsp;&nbsp;&nbsp;=\dfrac{95}{167} \\ \end{array}$

 There are four tube light slots in a room, four tube lights are placed in these slots out of a set of 9 tube lights. Out of these 9 only 3are in working condition. What is the probability that the room will be lit if power is supplied to all four slots?

 First find the probability of choosing four lights from nine, then find the probability if none of them worked, and finally subtract both to get the probability that any one will be lit.

$ &nbsp;&nbsp;&nbsp;\text{The room will be lit if at least one tube light is in working order}\text{.} \\ &nbsp;&nbsp;\text{Total number of ways of choosing four tube lights out of nine}={{\text{ }}^{9}}{{C}_{4}} \\ &nbsp;&nbsp;\text{The number of ways of choosing four tube lights such that none of them work } \\ &nbsp;&nbsp;{{\text{=}}^{6}}{{C}_{4}} \\ &nbsp;&nbsp;\text{Thus the number of ways of choosing four tube lights such that at least one of them works }={{\text{ }}^{9}}{{C}_{4}}{{-}^{6}}{{C}_{4}} \\ &nbsp;&nbsp;\text{The probability of room being lit }=\dfrac{^{9}{{C}_{4}}{{-}^{6}}{{C}_{4}}}{^{9}{{C}_{4}}}=\dfrac{\left( 126\text{ }-15 \right)}{126}=\dfrac{111}{126}=\dfrac{37}{42}. \\ $