Question

On an average, 20% of the computers in a firm are virus infected. If 10 computers are chosen at random from this firm, find the probability that at least one computer is virus infected, using Binomial distribution.

Answer 1

The probability that at least one computer is virus infected is $\frac{8717049}{9765625}$.

Answer 2

Let $n$ be the number of computers chosen, so $n=10$.

Let $p$ be the probability that a computer is virus infected, so $p = 20\% = 0.20$.

Let $q$ be the probability that a computer is not virus infected, so $q = 1 - p = 1 - 0.20 = 0.80$.

The number of virus infected computers among the 10 chosen follows a Binomial distribution $X \sim B(n, p)$, i.e., $X \sim B(10, 0.20)$.

The probability of getting exactly $k$ virus infected computers in 10 trials is given by the Binomial probability formula:

$P(X=k) = \binom{n}{k} p^k q^{n-k} = \binom{10}{k} (0.20)^k (0.80)^{10-k}$.

We need to find the probability that at least one computer is virus infected, which is $P(X \ge 1)$.

The event "at least one computer is virus infected" is the complement of the event "no computer is virus infected".

So, $P(X \ge 1) = 1 - P(X < 1) = 1 - P(X=0)$.

Now, we calculate $P(X=0)$:

$P(X=0) = \binom{10}{0} (0.20)^0 (0.80)^{10-0}$

$P(X=0) = 1 \cdot 1 \cdot (0.8)^{10}$

$P(X=0) = (0.8)^{10}$

We can write $0.8$ as a fraction: $0.8 = \frac{8}{10} = \frac{4}{5}$.

So, $P(X=0) = \left(\frac{4}{5}\right)^{10} = \frac{4^{10}}{5^{10}}$.

Calculate $4^{10}$ and $5^{10}$:

$4^{10} = (2^2)^{10} = 2^{20} = (2^{10})^2 = 1024^2 = 1048576$.

$5^{10} = (5^5)^2 = (3125)^2 = 9765625$.

So, $P(X=0) = \frac{1048576}{9765625}$.

Now, we calculate $P(X \ge 1)$:

$P(X \ge 1) = 1 - P(X=0)$

$P(X \ge 1) = 1 - \frac{1048576}{9765625}$

$P(X \ge 1) = \frac{9765625}{9765625} - \frac{1048576}{9765625}$

$P(X \ge 1) = \frac{9765625 - 1048576}{9765625}$

$P(X \ge 1) = \frac{8717049}{9765625}$

Alternatively, using decimal form:

$P(X=0) = (0.8)^{10} \approx 0.107374$

$P(X \ge 1) = 1 - 0.107374 = 0.892626$ (rounded to 6 decimal places).

The exact fractional answer is $\frac{8717049}{9765625}$.

Explanation of the solution:

The problem follows a Binomial distribution with $n=10$ trials (computers chosen) and probability of success $p=0.2$ (virus infected). The probability of failure is $q=1-p=0.8$. We want to find the probability of at least one success, $P(X \ge 1)$. This is calculated as $1 - P(X=0)$. Using the Binomial probability formula $P(X=k) = \binom{n}{k} p^k q^{n-k}$, we find $P(X=0) = \binom{10}{0} (0.2)^0 (0.8)^{10} = (0.8)^{10}$. Calculating $(0.8)^{10} = (4/5)^{10} = \frac{4^{10}}{5^{10}} = \frac{1048576}{9765625}$. The required probability is $1 - P(X=0) = 1 - \frac{1048576}{9765625} = \frac{9765625 - 1048576}{9765625} = \frac{8717049}{9765625}$.