Question

A bag contains 5 tennis balls and 4 cricket balls. Three balls are drawn at random from the bag. Find the probability that

1. all the drawn balls are cricket balls.

2. at least one of the drawn balls is a cricket ball.

Answer 1

Hint: In this type of question we have to find out the total number of outcomes and favorable cases using permutation and combinations. And then we will put the values in the given formula,

$P\left( E \right) = \dfrac{{{\text{number of favorable cases}}}}{{total{\text{ number of outcomes}}}}$

Complete step-by-step answer:

In this question, we are given that there is a bag containing 5 tennis balls and 4 cricket balls. And three balls are drawn at random.

So the total number of balls in the box are 9.

Now, we know that for any event E, probability of E is

$P\left( E \right) = \dfrac{{number{\text{ of ways E can occur}}}}{{total{\text{ number of outcomes}}}}$ -(1)

And here from 9 balls 3 balls are drawn so total number of outcomes are $\dfrac{{9!}}{{3!\left( {9 - 3} \right)!}} = \dfrac{{9!}}{{3!\left( 6 \right)!}} = \dfrac{{7 \times 8 \times 9}}{{2 \times 3}} = 7 \times 4 \times 3 = 84$  -- (2)

$\because $ total number of balls in bag is 9

Now according to question,

1. We have to find the probability when all the balls drawn are cricket balls.

So, let event A be such that all the three balls are cricket balls.

$P\left( A \right) = \dfrac{{{\text{number of ways A can occur}}}}{{{\text{total number of outcomes}}}}$ -(3)

Now number of ways A can occur means the number of ways by which all the three balls drawn are cricket balls is

$\dfrac{{4!}}{{3!\left( {4 - 3} \right)!}} = \dfrac{{4!}}{{3!\left( 1 \right)!}} = \dfrac{{4 \times 3 \times 2 \times 1}}{{3 \times 2 \times 1\left( 1 \right)}} = 4$  --(4)  $\because $ total number of cricket balls is 4

So, $P\left( A \right) = \dfrac{4}{{84}} = \dfrac{1}{{21}}$ - [using (2), (3) and (4)]

So, the required probability is $\dfrac{1}{{21}}$.

2. Now we have to find a probability when at least one of the balls is a cricket ball.

So now we have three cases,

First- 1 cricket ball and 2 tennis balls are drawn

Second- 1 tennis ball and 2 cricket balls are drawn

Third- 3 cricket balls are drawn

Let event L be such that 1 cricket ball and 2 tennis balls are drawn,

Event M be such that 1 tennis and 2 cricket balls are drawn and

Event N be such that 3 cricket balls are drawn.

Now let the event B be such that at least one of the balls drawn is cricket balls.

Now, required probability P(B) is

$P\left( B \right) = P\left( L \right) + P\left( M \right) + P\left( N \right)$ -(5)

Now, total number of outcomes is 84 so,

\[P\left( L \right) = \dfrac{{number{\text{ of ways L can occur}}}}{{total{\text{ number of outcomes}}}} = \dfrac{{\dfrac{{4!}}{{1!\left( {4 - 1} \right)!}}\dfrac{{5!}}{{2!\left( {5 - 2} \right)!}}}}{{84}}\]

\[{\text{ }} = \dfrac{{\dfrac{{4!}}{{1!\left( 3 \right)!}}\dfrac{{5!}}{{2!\left( 3 \right)!}}}}{{84}} \]

\[  {\text{ }} = \dfrac{{\dfrac{{4 \times 3 \times 2 \times 5 \times 4 \times 3 \times 2}}{{3 \times 2 \times 2 \times 3 \times 2}}}}{{84}} \]

\[{\text{ }} = \dfrac{{40}}{{84}} = \dfrac{{10}}{{21}}]\]

\[P\left( M \right) = \dfrac{{{\text{number of ways M can occur}}}}{{{\text{total number of outcomes}}}} = \dfrac{{\dfrac{{5!}}{{1!\left( {5 - 1} \right)!}}\dfrac{{4!}}{{2!\left( {4 - 2} \right)!}}}}{{84}} \]

\[{\text{ }} = \dfrac{{\dfrac{{5 \times 4 \times 3 \times 2}}{{1\left( 4 \right)!}}\dfrac{{4 \times 3 \times 2}}{{2\left( 2 \right)!}}}}{{84}} \]

\[{\text{ }} = \dfrac{{5 \times 6}}{{84}} = \dfrac{5}{{14}} \]

\[P\left( N \right) = \dfrac{{number{\text{ of ways N can occur}}}}{{{\text{total number of outcomes}}}} = \dfrac{{\dfrac{{4!}}{{3!\left( {4 - 3} \right)!}}}}{{84}} = \dfrac{1}{{21}}\]

now putting all these values in (5) we get,

$P(B) = \dfrac{{10}}{{21}} + \dfrac{5}{{14}} + \dfrac{1}{{21}} = \dfrac{{20 + 15 + 2}}{{42}} = \dfrac{{37}}{{42}}$

Hence the required probability is $\dfrac{{37}}{{42}}$.

Note: There is an alternative way to solve part 2,

As we have to find probability when at least one of the balls drawn is a cricket ball. Let p is the probability that no ball drawn is a cricket ball, which means all the balls drawn are tennis balls.

So, $P\left( {at{\text{ least one of the drawn balls is cricket ball}}} \right) = 1 - p$.

And here we can easily find value of p,

$p = \dfrac{{number{\text{ of favourable outcomes}}}}{{total{\text{ number of outcomes}}}} = \dfrac{{\dfrac{{5!}}{{3!\left( {5 - 3} \right)!}}}}{{84}} = \dfrac{{\dfrac{{5!}}{{3!\left( 2 \right)!}}}}{{84}} = \dfrac{{10}}{{84}} = \dfrac{5}{{42}}$

So, $P\left( {at{\text{ least one of the drawn ball is cricket ball}}} \right) = 1 - \dfrac{5}{{42}} = \dfrac{{37}}{{42}}$.