Question

There are two groups of subjects one of which consists of 5 science subjects and 3 engineering subjects and the other consist of 3 science and 5 engineering subjects. An unbiased die is cast. If number 3 or number 5 turns up, a subject is selected at random from the first group, otherwise the subject is selected at random from the second group. Find the probability that an engineering subject is selected ultimately.
 $
  (a)\dfrac{{13}}{{24}} \\
  (b)\dfrac{1}{8} \\
  (c)\dfrac{5}{{12}} \\
  (d)\dfrac{{11}}{{24}} \\
 $

Answer 1

Hint:In this question first of all find the probability of appearance of 3 or 5 while throwing up an unbiased dice using the basics of probability. That is probability is the ratio of the number of favorable outcomes to the total number of outcomes. Now find the probability to select an engineering subject from the first group and then the probability to select an engineering subject from the second group.

Complete step-by-step answer:
Given data:
First group: 5 science subjects and 3 engineering subjects, therefore in the first group there are total (3 + 5) = 8 subjects.
Second group: 3 science subjects and 5 engineering subjects, therefore in the second group there are total (5 + 3) = 8 subjects.
Now it is given that when an unbiased die is cast, if number 3 or 5 is shown then the first group is chosen, otherwise the second group is chosen.
Now as we know that in an unbiased dice there are a total six numbers (1, 2, 3, 4, 5 and 6).
So the probability of 3 or 5 shows up = (number of favorable outcomes/total number of outcomes) = (2/6) = (1/3).
Now as we know that the total probability is one (1), so the probability of the number 3 or 5 not showing up = 1 – (1/3) = (2/3).
Now the probability to select an engineering subject from the first group is = (number of favorable outcomes/total number of outcomes).
As there are 3 engineering subjects out of 8 in the first group so the favorable number of outcomes to select one engineering subject out of 3 is $ {}^3{C_1} $
And the total number of outcomes to select one engineering subject out of 8 for the first group is $ {}^8{C_1} $
So the probability to select an engineering subject from the first group is = \[\dfrac{{{}^3{C_1}}}{{{}^8{C_1}}}\]
Now the probability to select an engineering subject from the second group is = (number of favorable outcomes/total number of outcomes).
As there are 5 engineering subjects out of 8 in the second group so the favorable number of outcomes to select one engineering subject out of 5 is $ {}^5{C_1} $
And the total number of outcomes to select one engineering subject out of 8 for the second group is $ {}^8{C_1} $
So the probability to select an engineering subject from the second group is = \[\dfrac{{{}^5{C_1}}}{{{}^8{C_1}}}\]
Now the conditional probability that an engineering subject is selected is
P = probability to choose first bag $ \times $ probability to select an engineering subject from the first group + probability to choose second bag $ \times $ probability to select an engineering subject from the second group
Now substitute the values we have,
 $ \Rightarrow P = \dfrac{1}{3} \times \dfrac{{{}^3{C_1}}}{{{}^8{C_1}}} + \dfrac{2}{3} \times \dfrac{{{}^5{C_1}}}{{{}^8{C_1}}} $
Now simplify this according to property $ {}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}} $ we have,
 $ {}^3{C_1} = \dfrac{{3!}}{{1!\left( {3 - 1} \right)!}} = \dfrac{{3!}}{{2!}} = 3 $
Similarly
 $ {}^5{C_1} = 5 $
 $ {}^8{C_1} = 8 $
Therefore,
 $ \Rightarrow P = \dfrac{1}{3} \times \dfrac{3}{8} + \dfrac{2}{3} \times \dfrac{5}{8} = \dfrac{3}{{24}} + \dfrac{{10}}{{24}} = \dfrac{{13}}{{24}} $
So this is the required probability.
Hence option (A) is the correct answer.

Note: The trick concept used above is that since the probability of selecting an engineering subject is being asked and since engineering subject is in two group of subjects so eventually this ultimate probability will be the sum of the probability of getting an engineering subject from the first group of subject and the second group of subject but taking into consideration the fact that only the dice should show up 3 or 5.