Question

A lab network consisting of 20 computers was attacked by a computer virus. This virus enters each computer with probability 0.4, independently of other computers. Find the probability that it entered at least 10 computers

Answer 1

Title:

See the explanation.

Step-by-step explanation:

In the lab, there was a total of 20 computers. The probability of getting a computer affected because of the virus is 0.4.

Hence, the probability of a computer not getting affected is (1 - 0.4) = 0.6.

From the 20 computers, 10 can be chosen in $^{20}C_{10}$ ways.

The probability of 10 computers will be affected is $^{20}C_{10} \times (0.4)^{10} \times (0.6)^{10}$.

Now, the probability of getting 11 computers be affected is $^{20}C_{11}\times (0.4)^{11}\times(0.6)^9$.

Hence, the probability of at least 10 computers get affected is ∑$^{20}C_n \times (0.4)^n\times(0.6)^{10-n}$, where $10 \leq n \leq 20$.

Answer 2

Answer:

Probability that it entered at least 10 computers is 0.8725 .

Step-by-step explanation:

We are given that a lab network consisting of 20 computers was attacked by a computer virus. This virus enters each computer with probability 0.4, independently of other computers.

The above situation can be represented through Binomial distribution;

$P(X=x) = \binom{n}{x}p^{x} (1-p)^{n-x} ; x = 0,1,2,3,.....$

where, n = number of trials(samples) taken = 20 computers

           $x$ = number of success = at least 10

           p = probability of success which in our question is probability of

                  virus entering into each computer, i.e., 40%

LET X = Number of computers virus is entering into

So, it means X ~ $Binom(n= 20,p=0.40)$

Now, the probability that it entered at least 10 computers = P(X $\leq$ 10)

For calculating above probability we can use an easy method by using Binomial probability table which gives the value of P(X $\leq$ $x$);

So, P(X $\leq$ $x$) = P(X $\leq$ 10) = 0.8725

The above probability is calculated using binomial table where, n = 20, p = 0.4 and $x$ = 10 .