If six people decide to come to a basketball game, but three of them are only 2/5 sure that they will stay for the entire time (the other three are sure they'll stay the whole time), what is the probability that at the end, at least 5 people stayed the entire time

the probability that at the end, at least 5 people stayed the entire time = 0.352

Step-by-step explanation:

From the question, 3 of the people are sure to stay the whole time. So, we'll deduct 3 from 6.which leaves us with 3 that are only 2/5 or 0.4 sure that they will stay the whole time.

Thus, what we need to compute to fulfill the probability that at the end, at least 5 people stayed the entire time of which we know 3 will stay, so for the remaining 3,we'll compute;

P[≥2] which is x~bin(3,0.4)

Thus;

P(≥2) = (C(3,2) x 0.4² x 0.6) + (C(3,3) x 0.4³)

P(≥2) = 0.288 + 0.064

P(≥2) = 0.352

step-by-step explanation:

first, eliminate the parenthesis:

3

x

+

2

−

4

x

second, group like terms:

3

x

−

4

x

+

2

third, combine like terms:

(

3

−

4

)

x

+

2

→

−

1

x

+

2

→

−

x

+

2

  or  

step-by-step explanation:

if the four friends split the cost, p, of the pizza evenly, then each friend pays one-fourth of the cost of the pizza.

the expression is   or