Answer: The probability that she gets a ride 3 times in a 5-day work week is 0.31.
The probability that she gets a ride at least 2 times in a 5-day workweek is 0.97.
Step-by-step explanation:
By binomial distribution formula:
[tex]P[X=r]=^nC_rp^rq^{n-r}[/tex] , where n is the number of trials , r is the number of success, p is the probability of success and q is the probability of failure.
Given : n=5
p = 0.7
q= 1-0.7=0.3
Now, the probability that she gets a ride 3 times in a 5-day work week :
[tex]P[X=3]=^5C_3(0.7)^3(0.3)^{5-3}\\\\=10(0.7)^3(0.3)^2=0.3087\approx0.31[/tex]
The probability that she gets a ride at least 2 times in a 5-day workweek :
[tex]P[X\geq2]=1-P[X<2]\\\\=1-(P[X=1]+P[X=0])\\\\=1-(^5C_0(0.7)^0(0.3)^{5}+^5C_1(0.7)^1(0.3)^{5-1})\\\\=1-[(0.3)^5+5(0.7)(0.3)^4]\\\\=1-0.03078=0.96922\approx0.97[/tex]
Answer: that is wrong. It isn’t 0.31 and 0.97, because I just got it wrong.
Step-by-step explanation:
