Question

A bag contains 5 red and 3 blue balls. If 3 balls are drawn at random without replacement, the probability of getting exactly one red ball
(a) $${\displaystyle \frac{45}{196}}$$
(b) $${\displaystyle \frac{135}{392}}$$
(c) $${\displaystyle \frac{15}{56}}$$
(d) $${\displaystyle \frac{15}{29}}$$

Answer 1

Hint: Find the total number of balls in the bag. There are 3 cases, you might get one red ball either in the $$1^{st},2^{nd}$$ or the $$3^{rd}$$ draw. While 2 other balls are blue. Thus find the probability of these 3 cases and add them together.

Complete step-by-step answer:
It is said that a bag contains 5 red and 3 blue balls.
Thus the total number of balls in the bag = 5 red + 3 blue balls = 8 balls.
Now it is said that 3 balls are drawn out at random without replacement. Thus we need to get the probability of getting exactly one red ball and the other 2 balls as blue. Thus out of the 3 balls drawn one must be red and the other 2 as blue.
We know that the probability of an event is given as,
$Probability={\displaystyle \frac{numberoffavorableoutcomes}{Totalnumberofpossibleoutcomes}}$
Let us take P (R) as the probability of getting a red ball and P (B) as getting a blue ball.
Now we have three cases, we might get this red ball either in the first draw or second draw or third draw. Hence these three cases are possible.
Case 1: When we get a red in the first draw.
$$P=P\left(R\right)\times P{\left(B\right)}_1\times P{\left(B\right)}_2$$
P (R) = number of red balls / Total number of balls = $${\displaystyle \frac{5}{8}}$$
Now as the balls are drawn without any replacement, now the total number of balls = 7.
$$P{\left(B\right)}_1$$ = number of blue balls / total number of balls = $${\displaystyle \frac{3}{7}}$$.
$$P{\left(B\right)}_2$$ = number of blue balls / total number of balls = $${\displaystyle \frac{2}{6}}$$.
Because it is without any replacement the number of blue balls becomes 2 and the total number of balls becomes 6.
$$\begin{array}{rl}& \therefore P=P\left(R\right)\times P{\left(B\right)}_1\times P{\left(B\right)}_2\\ & P={\displaystyle \frac{5}{8}}\times {\displaystyle \frac{3}{7}}\times {\displaystyle \frac{2}{6}}={\displaystyle \frac{5}{56}}\end{array}$$
$$\therefore$$ When we draw a red ball first = $${\displaystyle \frac{5}{56}}$$ - (1)
Case 2: When we get a red in the second draw.
$$\begin{array}{rl}& P=P{\left(B\right)}_1\times P\left(R\right)\times P{\left(B\right)}_2\\ & P={\displaystyle \frac{3}{8}}\times {\displaystyle \frac{5}{7}}\times {\displaystyle \frac{2}{6}}={\displaystyle \frac{5}{56}}\end{array}$$
$$\therefore$$ When we get a red ball at the second draw = $${\displaystyle \frac{5}{56}}$$ - (2)
Case 3: When we get a red in the third draw.
$$\begin{array}{rl}& P=P{\left(B\right)}_1\times P{\left(B\right)}_2\times P\left(R\right)\\ & P={\displaystyle \frac{3}{8}}\times {\displaystyle \frac{2}{7}}\times {\displaystyle \frac{5}{6}}={\displaystyle \frac{5}{56}}\end{array}$$
$$\therefore$$ Getting a red ball at the $$3^{rd}$$ draw = $${\displaystyle \frac{5}{56}}$$ - (3)
Now let us add these 3 cases to get the total probability of getting exactly one red ball.
$$\therefore$$ Total probability = $${\displaystyle \frac{5}{56}}+{\displaystyle \frac{5}{56}}+{\displaystyle \frac{5}{56}}={\displaystyle \frac{15}{56}}$$.
$$\therefore$$ Thus the probability of getting exactly one red ball is $${\displaystyle \frac{15}{56}}$$.
$$\therefore$$ Option (c) is the correct answer.

Note: You can also calculate it together without so many steps as,
P (exactly one ball) = $$P\left(R\right)\times P\left(B\right)\times P\left(B\right)+P\left(B\right)\times P\left(R\right)\times P\left(B\right)+P\left(B\right)\times P\left(B\right)\times P\left(R\right)$$
P (exactly one ball) = $${\displaystyle \frac{5}{8}}\times {\displaystyle \frac{3}{7}}\times {\displaystyle \frac{2}{6}}+{\displaystyle \frac{3}{8}}\times {\displaystyle \frac{5}{7}}\times {\displaystyle \frac{2}{6}}+{\displaystyle \frac{3}{6}}\times {\displaystyle \frac{2}{7}}\times {\displaystyle \frac{5}{6}}$$
P (exactly one ball) = $${\displaystyle \frac{15}{56}}$$