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Hint: Find the probability that one bulb is not defective. Find the probability that none of the bulbs are defective from a single box. Use the formula for binomial distribution.

Complete step-by-step answer:

Let X be the random variable which denotes that a bulb is defective.

Let n be the number of boxes.

Let p be the probability that one bulb is defective.

Let q be the probability that one bulb is not defective.

Let r be the number of bulbs that are defective.

Given Data –

n = 10 and p = $\dfrac{1}{{50}}$

(Probability of p + Probability of q = 1, as they are two outcomes of a single event)

We get, q = 1 – p

= 1 – $\dfrac{1}{{50}}$

q = $\dfrac{{49}}{{50}}$

We need to find the probability that none of the bulbs are defective from a single box.

i.e. r = 0.

Here, the bulb can either be defective or not. These two are independent events. The occurrence of one event implies that the other event cannot occur.

Hence, the random variable follows a binomial distribution.

We calculate the probability of a binomial distribution using the formula

P(X) = \[{{\text{n}}_{{{\text{C}}_{\text{r}}}}}{{\text{p}}^{\text{r}}}{{\text{q}}^{{\text{n - r}}}}\]

Therefore, P (X = r) = P (X = 0) = \[{\text{1}}{{\text{0}}_{{{\text{C}}_0}}}{\left( {\dfrac{1}{{50}}} \right)^0}{\left( {\dfrac{{49}}{{50}}} \right)^{10}}\]

P = \[{\left( {\dfrac{{49}}{{50}}} \right)^{10}}\]

Hence the value of k =10.

Note: In such a problem the trick is to identify the distribution followed by the random variable. In this case it is a binomial distribution.

A binomial distribution is a specific probability distribution. It is used to model the probability of obtaining one of two outcomes, a certain number of times (r), out of a fixed number of trials (n) of a discrete random event.

A binomial distribution has only two outcomes: the expected outcome is called a success and any other outcome is a failure. The probability of a successful outcome is p and the probability of a failure is 1 - p.

Complete step-by-step answer:

Let X be the random variable which denotes that a bulb is defective.

Let n be the number of boxes.

Let p be the probability that one bulb is defective.

Let q be the probability that one bulb is not defective.

Let r be the number of bulbs that are defective.

Given Data –

n = 10 and p = $\dfrac{1}{{50}}$

(Probability of p + Probability of q = 1, as they are two outcomes of a single event)

We get, q = 1 – p

= 1 – $\dfrac{1}{{50}}$

q = $\dfrac{{49}}{{50}}$

We need to find the probability that none of the bulbs are defective from a single box.

i.e. r = 0.

Here, the bulb can either be defective or not. These two are independent events. The occurrence of one event implies that the other event cannot occur.

Hence, the random variable follows a binomial distribution.

We calculate the probability of a binomial distribution using the formula

P(X) = \[{{\text{n}}_{{{\text{C}}_{\text{r}}}}}{{\text{p}}^{\text{r}}}{{\text{q}}^{{\text{n - r}}}}\]

Therefore, P (X = r) = P (X = 0) = \[{\text{1}}{{\text{0}}_{{{\text{C}}_0}}}{\left( {\dfrac{1}{{50}}} \right)^0}{\left( {\dfrac{{49}}{{50}}} \right)^{10}}\]

P = \[{\left( {\dfrac{{49}}{{50}}} \right)^{10}}\]

Hence the value of k =10.

Note: In such a problem the trick is to identify the distribution followed by the random variable. In this case it is a binomial distribution.

A binomial distribution is a specific probability distribution. It is used to model the probability of obtaining one of two outcomes, a certain number of times (r), out of a fixed number of trials (n) of a discrete random event.

A binomial distribution has only two outcomes: the expected outcome is called a success and any other outcome is a failure. The probability of a successful outcome is p and the probability of a failure is 1 - p.

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