Question

A bag has 4 red and 5 black balls and a second bag has 3 red and 7 black balls. One ball is drawn from the first and two from the second. Find the probability that out of three balls two are black and one is red.

Answer 1

Hint - We will start solving this question by writing down all the information mentioned in the question. Then by using the formula of probability of an event, i.e., Probability = Number of favorable outcomes / Total number of outcomes, we will find the required solution.

Complete step-by-step solution -
It is given that there are two bags, bag A and bag B. In bag A there are 4 red balls and 5 black balls and in bag B there are 3 red balls and 7 black balls. One ball is drawn from bag A and two balls are drawn from bag B.

We have to find the probability of getting one red ball and two black balls.

Let $E$ be the event of getting one red ball and two black balls, then probability to be found is $P\left( E \right)$.

Now, the probability $P\left( E \right)$can be found by dividing this event into three different events, i.e., ${E_1},{E_2}$and ${E_3}$.

In first event, i.e., ${E_1}$, when we draw a ball from bag A let it be a black ball, then its probability is $\dfrac{5}{9}$, as there are a total of 9 balls in bag A of which 5 are black and when we draw two balls from bag B let the first ball be black, then its probability is $\dfrac{7}{{10}}$, as there are a total of 10 balls in bag B of which 7 are black and the second ball be red, then its probability is $\dfrac{3}{9}$, as after drawing one ball there are 9 balls left in the bag of which 3 are red. Then the total probability of this event would be,
$P\left( {{E_1}} \right)$ $ = \dfrac{5}{9} \times \dfrac{7}{{10}} \times \dfrac{3}{9}$
$
   = \dfrac{5}{9} \times \dfrac{7}{{10}} \times \dfrac{1}{3} \\
   = \dfrac{{35}}{{270}} \\
 $
 $
   = \dfrac{5}{9} \times \dfrac{7}{{10}} \times \dfrac{1}{3} \\
   = \dfrac{{35}}{{270}} \\
 $

In second event, i.e., ${E_2}$, when we draw a ball from bag A let it be a black ball, then its probability is $\dfrac{5}{9}$, as there are a total of 9 balls in bag A of which 5 are black and when we draw two balls from bag B let the first ball be red, then its probability is $\dfrac{3}{{10}}$, as there are a total of 10 balls in bag B of which 3 are red and the second ball be black, then its probability is $\dfrac{7}{9}$ , as after drawing one ball there are 9 balls left in the bag of which 7 are red. Then the total probability of this event would be,
$P\left( {{E_2}} \right) = \dfrac{5}{9} \times \dfrac{3}{{10}} \times \dfrac{7}{9}$
 $
   = \dfrac{5}{9} \times \dfrac{1}{{10}} \times \dfrac{7}{3} \\
   = \dfrac{{35}}{{270}} \\
 $

In third event, i.e., ${E_3}$, when we draw a ball from bag A let it be a red ball, then its probability is $\dfrac{4}{9}$, as there are a total of 9 balls in bag A of which 4 are red and when we draw two balls from bag B let the first ball be black, then its probability is $\dfrac{7}{{10}}$, as there are a total of 10 balls in bag B of which 7 are black and the second ball is also black, then its probability is $\dfrac{6}{9}$ , as after drawing one ball there are 9 balls left in the bag of which 6 are black, as one is already drawn. Then the total probability of this event would be,
$P\left( {{E_3}} \right) = \dfrac{4}{9} \times \dfrac{7}{{10}} \times \dfrac{6}{9}$
 $
   = \dfrac{4}{9} \times \dfrac{7}{{10}} \times \dfrac{2}{3} \\
   = \dfrac{{56}}{{270}} \\
 $

Therefore, total probability $P\left( E \right)$is,
$P\left( E \right)$ $ = P\left( {{E_1}} \right) + P\left( {{E_2}} \right) + P\left( {{E_3}} \right)$
   $
   = \dfrac{{35}}{{270}} + \dfrac{{35}}{{270}} + \dfrac{{56}}{{270}} \\
   = \dfrac{{126}}{{270}} \\
   = \dfrac{{42}}{{90}} \\
   = \dfrac{{14}}{{30}} \\
   = \dfrac{7}{{15}} \\
 $

Hence, the probability of getting two are black and one is red is $\dfrac{7}{{15}}$.

Note- The probability of an event is the likelihood of that event occurring. It is a value between and including zero and one. These kinds of questions are very simple if one knows the basic formulas and the formulas must be remembered as there is no alternate method for solving this question.