Question

The probability that an electronic device produced by a company does not function properly is equal to $0.1$. If 10 devices are bought, then the probability, to the nearest thousandth, than 7 devices function properly is
A.$0.057$
B.$0.478$
C.$0.001$
D.0

Answer 1

Here, we will find the probability that 7 devices function properly by using the Binomial distribution. We will find the probability that an electronic device produced by a company function properly using the formula sum of an event and a complementary event. We will then substitute values of all the probabilities in the formula of Binomial Distribution to find the required probability. Binomial Distribution is the probability distribution of the number of successes in a sequence of independent trials.

Formula Used:
We will use the following formulas:
1.Sum of an event and a complementary event is given by $p + q = 1$
2.Binomial Probability Distribution is given by the formula ${P_r} = {}^n{C_x}{p^x}{q^{n - x}}$ , where $P$ is the binomial Probability, $x$ is the number of times that the outcome occurs in n-trials, $n$ is the number of trials, ${}^n{C_x}$ is the number of combinations, $p$ is the probability of success, $q$ is the probability of failures.
3.Combination is given by the formula ${}^n{C_r} = \dfrac{{n!}}{{(n - r)!r!}}$ .

Complete step-by-step answer:
Let $p$ be the probability that an electronic devices function properly and $q$ be the probability that an electronic devices that does not function properly respectively.
We are given that the probability that an electronic device produced by a company does not function properly is equal to $0.1$.
So, $q = 0.1$
Thus, we can find the probability that an electronic device produced by a company function properly by sum of an event and a complementary event. So,
$p + q = 1$
Substituting $q = 0.1$ in the above equation, we get
$\Rightarrow p = 1 - 0.1 = 0.9$
Since, we got the probability of success or failure, so we will use the binomial probability distribution.
Substituting $n = 10,x = 7,p = 0.9$ and $q = 0.1$ in the formula ${P_r} = {}^n{C_x}{p^x}{q^{n - x}}$, we get
$\Rightarrow {P_7} = {}^{10}{C_7} \times {(0.9)^7} \times {(0.1)^{10 - 7}}$
Subtracting the term in the exponent, we get
$\Rightarrow {P_7} = {}^{10}{C_7} \times {(0.9)^7} \times {(0.1)^3}$
Simplifying the equation, we get
$\Rightarrow {P_7} = {}^{10}{C_7} \times 0.4782 \times 0.001$
Multiplying the terms, we get
$\Rightarrow {P_7} = {}^{10}{C_7} \times 0.0004782$
Now by using the formula of Combination ${}^n{C_r} = \dfrac{{n!}}{{(n - r)!r!}}$, we get
$\Rightarrow {P_7} = \dfrac{{10!}}{{(10 - 7)!7!}} \times 0.0004782$
Subtracting the terms in the bracket, we get
$\Rightarrow {P_7} = \dfrac{{10!}}{{3!7!}} \times 0.0004782$
Computing the factorial, we get
$\Rightarrow {P_7} = 10 \times 3 \times 4 \times 0.0004782$
Multiplying the terms, we get
$\Rightarrow {P_7} = 0.057$
Therefore, the probability that the electronic devices function properly is $0.057$.
Hence, option A is the correct answer.

Note: Here, we are using the binomial distribution because we have only two outcomes either success or failure. Binomial distribution should satisfy the conditions that include the number of trials is constant. Each observation is independent from all the other trials since it has no effect on the other. The probability of success is the same for all the trials. We should know that the binomial distribution should be used only if it satisfies all the conditions.