Hi! This problem follows the binomial distribution where there are only two outcomes - either Sandra gets a ride to work or not. A binomial distribution is modeled by the following equation:
[tex]P(x)= _{n}C_{x} p^{x} q^{n-x} [/tex]
where p is the probability of success for x trials and q is the probability of failure (n is the total number of trials). In the case of the problem p would be the probability of getting a ride which is equal to 0.7 while q would be 0.3.
To find the probability that she successfully gets a ride 3 out of 5 times, we just substitute 3 as the value of x:
[tex]P(3)= _{5}C_{3} (0.7)^{3} (0.3)^{2} =0.3087[/tex]
Meanwhile, to find the probability that she gets AT LEAST 2 out of 5 rides, we just get the probability that she gets 1 ride or no ride at all then subtract the sum of these two probabilities from 1.
[tex]P(0)= _{5}C_{0} (0.7)^{0} (0.3)^{5} =0.00243[/tex]
[tex]P(1)= _{5}C_{1} (0.7)^{1} (0.3)^{4} =0.02835[/tex]
[tex]P(x \geq 2)=1-[P(0)+P(1)]=1-(0.00243+0.02835)=1-0.03078[/tex]
[tex]P(x \geq 2)=0.96922[/tex]
ANSWER: The probability that Sandra gets a ride 3 times in a 5-day work is 0.309, and the probability that she gets a ride at least 2 times in a 5-day work is 0.969.
