prob(A) = 2/x
prob(B) = 2/x
prob(C) = 1/x
but 2/x + 2/x + 1/x = 1
times x
5 = x
A--> 2/5
B--> 2/5
C--> 1/5
order of finish
ABC *** (2/5)(3/5)(4/5) = 24/125
ACB *** (2/5)(4/5)(3/5) = 24/125
BAC ---
BCA ---
CAB ---
CBA ---
We don't want the two I calculated, so
prob(B or C) = 1 - 48/125 = 77/125
Do a similar analysis for #2
3 students; A, B and C, are in a swimming race. A and B have the same probability of winning and each is twice as likely to win as C. Find the probability that B or C wins the swimming race.
3) Given three boxes as follows:
Box A contains 3 red, 2 white and 2 blue balls.
Box B contains 2 red, 2 white and 3 blue balls.
Box C contains 2 red, 3 white and 2 blue balls.
First, a box is selected at random and than two balls are selected from that box.
a) What is the probability that the selected both balls are white?
b) If the selected both balls are white, what is the probability that they are selected from Box B?
c) What is the probability that the selected balls are of the same colour?
1 answer