Question

A box in a machine shop consists of 5 coated and 10 uncoated cutting inserts which are of otherwise similar characteristics. The inserts got mixed up randomly in the box. An operator has taken 4 of them at once without noticing the differences to mount on a 4-tooth face milling cutter that uses inserts.

The probability of the milling cutter having all uncoated inserts is ______ (rounded off to two decimal places).

Hint:

Use the multiplication rule for drawing without replacement: \( \frac{10}{15} \times \frac{9}{14} \times \frac{8}{13} \times \frac{7}{12} \), or equivalently the ratio of combinations \( \frac{^{10}C_4}{^{15}C_4} \).

Answer 1

Correct Answer: 0.15

Step 1: Understanding the Concept:
The box has 15 cutting inserts in total, 5 coated and 10 uncoated, and they are mixed randomly.
The operator picks up 4 inserts at once, so we are picking without replacement from this batch.
We need the probability that all 4 picked inserts turn out to be uncoated.

Step 2: Key Formula or Approach:
Since the inserts are drawn one after another without putting any back, we can find the required probability by multiplying the probability of each successive draw being uncoated.
This is the same idea as combinations, just written as a chain of conditional probabilities.
\[ P(\text{all 4 uncoated}) = \frac{10}{15} \times \frac{9}{14} \times \frac{8}{13} \times \frac{7}{12} \]

Step 3: Detailed Explanation:
On the first draw, 10 of the 15 inserts are uncoated, so the chance is \( \frac{10}{15} \).
Once one uncoated insert is removed, 9 uncoated inserts remain out of 14 total, giving \( \frac{9}{14} \).
Continuing this way, the third draw gives \( \frac{8}{13} \) and the fourth draw gives \( \frac{7}{12} \).
Multiplying these together:
\[ P = \frac{10 \times 9 \times 8 \times 7}{15 \times 14 \times 13 \times 12} = \frac{5040}{32760} \]
\[ P = 0.15384... \approx 0.15 \]

Final Answer:
Rounded to two decimal places, the probability that the milling cutter has all uncoated inserts is 0.15. \[ \boxed{0.15} \]