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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).

Probability Calculation for Uncoated Inserts

This problem involves calculating the probability of a specific outcome when selecting items randomly from a group. We need to find the chance that all 4 selected cutting inserts are uncoated.

Understanding the Problem

  • Total number of cutting inserts = 5 (coated) + 10 (uncoated) = 15.
  • Number of inserts selected = 4.
  • We need the probability that all 4 selected inserts are from the 10 uncoated ones.

Step 1: Calculate Total Possible Selections

The total number of ways to choose any 4 inserts from the 15 available is calculated using combinations ($C(n, k)$), where $n$ is the total number of items and $k$ is the number of items to choose.

Total combinations = $C(15, 4) = \frac{15!}{4!(15-4)!} = \frac{15!}{4!11!}$

Calculation: $\frac{15 \times 14 \times 13 \times 12}{4 \times 3 \times 2 \times 1} = 1365$ ways.

Step 2: Calculate Favorable Selections

The number of ways to choose 4 uncoated inserts from the 10 available uncoated inserts is also calculated using combinations.

Favorable combinations = $C(10, 4) = \frac{10!}{4!(10-4)!} = \frac{10!}{4!6!}$

Calculation: $\frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} = 210$ ways.

Step 3: Determine the Probability

The probability is the ratio of favorable selections to the total possible selections.

Probability (all 4 uncoated) = $\frac{\text{Favorable combinations}}{\text{Total combinations}} = \frac{C(10, 4)}{C(15, 4)}$

Probability = $\frac{210}{1365}$

Step 4: Final Calculation and Rounding

Simplify the fraction and convert to a decimal:

Probability = $\frac{210}{1365} = \frac{2}{13} \approx 0.1538...$

Rounding to two decimal places, the probability is 0.15.

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Important Questions from Probability

  1. If one event is unaffected by the outcome of another event, the two events are said to be:
    1. Dependent
    2. Independent
    3. Mutually exclusive
    4. Mutually inclusive
  2. Let $M$ be a randomly chosen non-empty subset of $S = \{1, 2, 3, \dots, 2026\}$.

    Which of the following is the probability that the product of all the elements of $M$ is even?
  3. A fair coin is tossed 6 times. The probability of getting at least 5 heads is _________________. (rounded off to two decimal places)
  4. Consider a random experiment of rolling a fair die. The probability of getting an odd number or a number less than 4 is ______ (Rounded off to two decimal places).
  5. In a null hypothesis significance test, the significance level $\alpha$ was set to 0.01. For which one or more of the following $p$ values would the null hypothesis be rejected?
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