A box holds 15 items, of which 3 are defective and 12 are good. Five items are drawn at random without replacement. Find (1) the probability that exactly 2 are defective, and (2) the probability that at least 2 are defective.

Step-by-step solution

1. 1

   Set up the sample space as unordered selections. Choosing $5$ of the $15$ items without regard to order gives

   $\binom{15}{5}=\frac{15\cdot 14\cdot 13\cdot 12\cdot 11}{5!}=3003$

   equally likely outcomes. Because the draw is without replacement, combinations — not powers — are the right count.

2. 2

   (1) Count the favourable selections. Exactly two defectives means choosing $2$ of the $3$ defectives and $3$ of the $12$ good items:

   $\binom{3}{2}\binom{12}{3}=3\times 220=660$

3. 3

   Divide and reduce.

   $P(\text{exactly }2)=\frac{660}{3003}=\frac{20}{91}\approx 0.2198$

   (Both numbers are divisible by $33$.)

4. 4

   (2) Enumerate the cases for "at least 2". Since the box contains only $3$ defectives, "at least $2$" means exactly $2$ or exactly $3$ — just two cases, so no complement is needed:

   $\binom{3}{3}\binom{12}{2}=1\times 66=66$

5. 5

   Add the two disjoint cases.

   $P(\text{at least }2)=\frac{660+66}{3003}=\frac{726}{3003}=\frac{22}{91}\approx 0.2418$

   $\boxed{P(\text{exactly }2)=\tfrac{20}{91},\qquad P(\text{at least }2)=\tfrac{22}{91}}$

6. 6

   Check that the full distribution sums to 1. The four possible defect counts give $\binom{3}{0}\binom{12}{5}=792$, $\binom{3}{1}\binom{12}{4}=1485$, $660$ and $66$. Their total is $792+1485+660+66=3003=\binom{15}{5}$ ✓, which confirms every count above.