Question

A box contains 6 white, 7 purple and 3 black balls. Balls are picked randomly one at a time without replacement. Prob. (picking purple then black) is.

• A. $\frac{21}{256}$
• B. $\frac{3}{16}$
• C. $\frac{7}{80}$
• D. $\frac{7}{16}$

Answer 1

The correct option is C $\frac{7}{80}$

In the box,
number of white balls= $6$
number of purble balls= $7$
number of black balls= $3$
$\therefore$ Total number of balls in the box= $6+7+3=16$

Now,
$\text{P(picking a purple ball)}=\frac{\text{Number of purple balls}}{\text{Total number of balls}}$

$\Rightarrow \text{P(picking a purple ball)}=\frac{7}{16}$

After picking a purple ball without replacement,
number of white balls= $6$
number of purple balls= $7-1=6$
number of black balls= $3$
$\therefore$ Currently, total number of balls in the box= $6+6+3=15$

Now,
$\text{P(picking a black ball)}=\frac{\text{Number of black balls}}{\text{Total number of balls}}$

$\Rightarrow \text{P(picking a black ball)}=\frac{3}{15}=\frac{1}{5}$

Finally,
$\text{P(picking a purple then a black ball)= P(picking a purple ball)}\times \text{P(picking a black ball)}$

$\Rightarrow \text{P(picking a purple then a black ball)}=\frac{7}{16}\times \frac{1}{5}$

$\Rightarrow \text{P(picking a purple then a black ball)}=\frac{7\times 1}{16\times 5}=\frac{7}{80}$

$\therefore$ The probability of picking a purple then a black ball without replacement is $\frac{7}{80}$.