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?
To find the probability that B or C wins the swimming race, let's first determine the individual probabilities of each student winning.
Let P(A) be the probability that student A wins.
Let P(B) be the probability that student B wins.
Let P(C) be the probability that student C wins.
Given that A and B have the same probability of winning and each is twice as likely to win as C, we can write the following equations:
P(A) = 2P(C)
P(B) = 2P(C)
The sum of the probabilities of all three students winning must be 1:
P(A) + P(B) + P(C) = 1
Substituting the equations above into this sum equation, we get:
2P(C) + 2P(C) + P(C) = 1
5P(C) = 1
P(C) = 1/5
Since P(A) = 2P(C), we have P(A) = 2/5.
Similarly, P(B) = 2/5.
To find the probability that B or C wins, we add the probabilities of B and C winning:
P(B or C) = P(B) + P(C) = 2/5 + 1/5 = 3/5
Therefore, the probability that B or C wins the swimming race is 3/5.
Now let's move on to the second question about the boxes and ball selection.
a) What is the probability that the selected both balls are white?
To find the probability of selecting two white balls, we need to consider all possible combinations of selecting two balls from each box.
The total number of balls in Box A is 3 + 2 + 2 = 7.
The total number of balls in Box B is 2 + 2 + 3 = 7.
The total number of balls in Box C is 2 + 3 + 2 = 7.
The total number of balls across all three boxes is 7 + 7 + 7 = 21.
The number of ways to select two white balls from Box A is C(2, 7) = 7! / (2! * (7 - 2)!) = 21.
Similarly, the number of ways to select two white balls from Box B is C(2, 7) = 21.
And the number of ways to select two white balls from Box C is C(2, 7) = 21.
The total number of ways to select two white balls from all three boxes is 21 + 21 + 21 = 63.
Therefore, the probability of selecting two white balls is 63 / 21 = 3/7.
b) If the selected both balls are white, what is the probability that they are selected from Box B?
To find this conditional probability, we need to find the probability of selecting two white balls from Box B and divide it by the probability of selecting two white balls from any of the three boxes.
The probability of selecting two white balls from Box B is 21 / 63 = 1/3 (as calculated in part a).
The probability of selecting two white balls from any of the three boxes is 3/7 (as calculated in part a).
Therefore, the probability that the two white balls are selected from Box B given that both balls are white is (1/3) / (3/7) = 7/9.
c) What is the probability that the selected balls are of the same color?
To find the probability that the selected balls are of the same color, we need to consider all possible combinations of selecting two balls of the same color from each box.
The number of ways to select two balls of the same color from Box A is C(2, 7) + C(2, 3) = 21 + 3 = 24.
Similarly, the number of ways to select two balls of the same color from Box B is C(2, 7) + C(2, 3) = 24.
And the number of ways to select two balls of the same color from Box C is C(2, 7) + C(2, 3) = 24.
The total number of ways to select two balls of the same color from all three boxes is 24 + 24 + 24 = 72.
The total number of ways to select two balls from all three boxes is C(2, 21) = 21! / (2! * (21 - 2)!) = 210.
Therefore, the probability that the selected balls are of the same color is 72 / 210 = 12/35.
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