Question

A lot contains $10$ items out of which $3$ are defective. Three items are chosen from the lot one after another at random without replacement. Find the probability that all three are defective.

Answer 1

Step $\mathbf{1}$ : For the first draw:

Total number of defective balls $=3$

Total number of balls in the bag $=10$

Probability of drawing a defective ball $=\frac{totalnumberordefectiveballs}{totalnumberofballsinthebag}$

$=\frac{3}{10}$

Step $\mathbf{2}$ : For the second draw:

Since, the balls have to be picked without replacement.

So, number of remaining defective balls $=2$

And, total number of remaining balls in the bag $=9$

Therefore, probability of drawing defective ball $=\frac{totalnumberordefectiveballs}{totalnumberofballsinthebag}$

$=\frac{2}{9}$

Step $\mathbf{3}$ : For the third draw:

Now, the number of defective balls $=1$

And, total number of balls in the bag $=8$

And, probability of drawing defective ball $=\frac{totalnumberordefectiveballs}{totalnumberofballsinthebag}$

$=\frac{1}{8}$

Final Result: Probability that all three balls are defective:

Probability of drawing all three defective balls $=\frac{3}{10}\times \frac{2}{9}\times \frac{1}{8}$

$=\frac{\mathbf{1}}{\mathbf{120}}$

Hence, the probability that all three balls are defective is $\frac{\mathbf{1}}{\mathbf{120}}$.