Question

In a case of 144 boxes of light bulbs, it is found that exactly 4 boxes contain at least


one broken light bulb. If 5 boxes are selected from the case at random, what is the

probability (rounded to 4 decimal places) that none of the boxes will contain a broken

light bulb?

Answer 1

Answer on Question #42067 - Math - Statistics and Probability

In a case of 144 boxes of light bulbs, it is found that exactly 4 boxes contain at least one broken light bulb. If 5 boxes are selected from the case at random, what is the probability (rounded to 4 decimal places) that none of the boxes will contain a broken light bulb?

Solution

For a start, we number the boxes. Let, boxes with numbers from 1 to 4 contain at least one broken light bulb and boxes with number from 5 to 144 don't contain broken bulbs.

Let $\Omega = \{(\omega_1, \omega_2, \dots, \omega_5) | \omega_i = \overline{1,144}, \omega_i \neq \omega_j, if i \neq j\}$ – probability space. $|\Omega| = C_{144}^5$ , where $|\Omega|$ is a power of the probability space.

Now, we find the probability that none of the boxes will contain a broken light bulb. We denote it $P(A)$ .

$A = \{(\omega_{1},\omega_{2},\dots,\omega_{5})\in \Omega |\omega_{i} = \overline{5,144},\omega_{i}\neq \omega_{j},if i\neq j\} ,|A| = C_{140}^{5}$ , where $|A|$ is a power of $A$

Using classical definition of probability, we have

$$P (A) = \frac {| A |}{| \Omega |} = \frac {C _ {1 4 0} ^ {5}}{C _ {1 4 4} ^ {5}} = \frac {1 4 0 !}{5 ! \cdot 1 3 5 !} \cdot \frac {5 ! \cdot 1 3 9 !}{1 4 4 !} = \frac {1 3 6 \cdot 1 3 7 \cdot 1 3 8 \cdot 1 3 9}{1 4 1 \cdot 1 4 2 \cdot 1 4 3 \cdot 1 4 4} = 0. 8 6 6 9$$

**Answer**: the probability (rounded to 4 decimal places) that none of the boxes will contain a broken light bulb is 0.8669.

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