A school newspaper reporter decides to randomly survey 15 students to see if they will attend Tet festivities this year. Based on past years, she knows that 24% of students attend Tet festivities. We are interested in the number of students who will attend the festivities. Find the probability that at most 4 students will attend. (Round your answer to four decimal places.)

0.70319018

Step-by-step explanation:

Given the following:

Number of students surveyed (n) = 15

Probability of attending tet festival (p) = 24% =0.24

Therefore,

Probability of not attending (1 - p) = (1 - 0.24) = 0.76.

The probability that at most 4 students will attend can be obtained using the binomial probability relation:

p(x) = nCx * p^x * (1 - p)^(n-x)

At most 4 students means:

p(x=0) + p(x=1) + p(x=2) + p(x=3) + p(x=4)

p(x=0) = 15C0 * 0.24^0 * 0.76^(15 - 0)

p(x=0) = 1 * 1 * 0.0004701 = 0.00047018

p(x=1) = 15C1 * 0.24^1 * 0.76^(14)

p(x=1) = 15 * 0.24 * 0.021448 = 0.07721

p(x=2) = 15C2 * 0.24^2 * 0.76^(13) =

p(x=2) = 105 * 0.0576 * 0.02822 = 0.17068

p(x=3) = 15C3 * 0.24^3 * 0.76^(12)

p(x=3) = 455 * 0.013824 * 0.037133 = 0.23356

p(x=4) = 15C4 * 0.24^4 * 0.76^(11) =

p(x=4) = 1365 * 0.0033177 * 0.048859 = 0.22127

0.00047018 + 0.07721 + 0.17068 + 0.23356 + 0.22127 = 0.70319018

idk

step-by-step explanation: