A certain university has 10 vehicles available for use by faculty and staff. four of these are vans and 6 are cars. on a particu
lar day, only two requests for vehicles have been made. suppose that the two vehicles to be assigned are chosen in a completely random fashion from among the 10. (a) let e denote the event that the first vehicle assigned is a van. what is p(e)? 0.4 correct: your answer is correct. (b) let f denote the probability that the second vehicle assigned is a van. what is p(f | e)? incorrect: your answer is incorrect. (c) use the results of parts (a) and (b) to calculate p(e and f) (hint: use the definition of p(f | e).)
<span>There are 4 vans. So we have that probability that the first vehicle is a van p (e) = 4/10 = 0.4.
P(e|f) = P( f and e) / p (f)
p(e) = 0.4 and p(f) = 3/9
P (f and e) = 0.40 * 0.33 = 0.132
So p(e|f) = 0.4 * 0.33/ 0.33 = 0.4
P(f and e) p(f) * p(e) = 0.4 * 0.33 = 0.132</span>
9, the tens digit, is six more than 3, the ones digit. 1, the hundreds digit, is 8 less than the tens digit, 9. It can't be any other number combination than this.
