Question

A box has $2$ blue and $3$ yellow buttons of the same size. Two buttons are chosen at random without replacement. What is the probability that both are blue?

• A. $\frac{2}{5}$
• B. $\frac{1}{5}$
• C. $\frac{1}{10}$
• D. None of the above

Answer 1

The correct option is C $\frac{1}{10}$

Suppose, $\text{A}$ is an event of firstly drawing a blue botton and $\text{B}$ is an event of secondly drawing a blue botton.

As the second button is drawn $\text{without replacing}$ the first button, the secondly picked color depends on the firstly picked color button. Hence, the events $\text{A}$ and $\text{B}$ are $\text{dependent events}$.

$\therefore \text{P(A and B)=P(A)}\times \text{P(B knowing that A already occured)}$

In the given box,
number of blue buttons= $2$
number of yellow buttons= $3$
$\therefore$ Total number of buttons= $2+3=5$

Now,
$\text{P(A)}=\frac{\text{Number of blue bottons in the box}}{\text{Total number of buttons in the box}}$

$\Rightarrow \text{P(A)}=\frac{2}{5}$

Next,
$\text{P(B knowing that A already occured)}=\frac{\text{Number of blue buttons remaining in the box}}{\text{Total number of buttons remaining in the box}}$

$\Rightarrow \text{P(B knowing that A already occured)}=\frac{2-1}{5-1}=\frac{1}{4}$

Finally,
$\text{P(A and B)=P(A)}\times \text{P(B knowing that A already occured)}$

$\Rightarrow \text{P(A and B)}=\frac{2}{5}\times \frac{1}{4}$

$\Rightarrow \text{P(A and B)}=\frac{2\times 1}{5\times 2\times 2}=\frac{1}{10}$

$\therefore$ The probability of choosing both blue buttons is $\frac{1}{10}$.