Question

When Julia is writing a first draft, there is 0.7 probability that there will be no spelling mistakes on a page. One day, Julia writes a first draft that is 4 pages long. Assuming that Julia is equally likely to have a spelling mistake on each of the 4 pages, what is the probability that she will have no spelling mistakes on at least one of them?

Answer 1

Answer:

The required probability is  given by, 0.9919.

Step-by-step explanation:

Let, X be the random variable denoting the no. of pages among those 4 pages which Julia writes where she makes no spelling mistake.

clearly,

X $\sim$ Binomial (4, 0.7)

So,      P(X = x) = $^4C_{x} \times (0.7)^{x} \times (0.3)^{(4 - x)}$

                            [when  x = 0, 1, 2, 3, 4]

                         = 0 otherwise

According to the question, we are to find out  P(X ≥ 1) .

Now, P(X ≥ 1)

     = 1 - P(X = 0)

     = $1 - (^4C_{0} \times (0.7)^{0} \times (0.3)^{4})$

     = $1 - 0.0081$

     = 0.9919

So, the required probability is  given by, 0.9919

Answer 2

Answer:

real answer is 0.9919 but it depends if needed to round it is .99

Step-by-step explanation:

it was khan academy algebra 2 assignement