If the probability of a succesful log in is $0.8$, then what is the probability that there will be less than five succesful attempts in ten tries?

Question

The probability of being able to log on to a computer system from a remote terminal during a busy working day is $0.8$. Suppose that $10$ independent attempts are made.
What is the probability that
(a) less than five attempts will be successful?
(b) all attempts will be successful?

My attempt:
(a) I'm not sure about this one but I did, probability of success of five attempts - probability of failure so $.8^5 - .2^{10} = .32767$ (Is that right?)

(b) Each attempt's success is $.8$. Since there are $10$ attempts, probability will be $.8^{10} = .10737$.

Can someone tell me if I did these correctly? Thank you.

Answer 1

Notice that you have $10$ independent events with probability $p =.8$ of success. If we call the number of successful attempts $X$, then $X$ follows a binomial distribution, $$X\sim\text{Bin}(10, .8).$$

a) This ask for $P(X<5) = P(X\leq 4)$. Since each event is disjoint, we have that \begin{align*} P(X\leq 4) &=P(X=0)+P(X=1)+\dotsb+P(X = 4)\\ &= \sum_{k = 0}^4\binom{10}{k}(.8)^k(1-.8)^{10-k} \\ &= 0.006369382. \end{align*}

b) This is fine. If we frame this in terms of $X$, then $$P(X =10) = \binom{10}{10}(.8)^{10}(1-.8)^0 = 0.1073742.$$

Answer 2

Part (b) is correct, part (a) is not.

Hint:

You must have studied the binomial distribution to get this question.

Use it to compute the probability of $0$ successes, $1$ success, ..... $4$ successes, and add up

Answer 3

This is the binomial distribution.

$X \sim B(n,p)$

The binomial distribution with parameters $n$ and $p$ is the probability distribution of the number of successes in a sequence of $n$ independent trials, each of which yields success with probability $p$.

$a)$

$$P(X \leq 4) = \sum_{k=0}^{4} \binom{10}{k} 0.8^k(0.2)^{10-k}$$