I know I posted this already but no one answered the questions!!!!!
**************ANSWER THIS PLEASE!!!!
Two students are chosen at random from all 50 students.
Find the probability that
a) Both students walked
b) only one of the students walked
50 students total.
12 students travelled by car only
7 students travelled by bus only
5 students travelled by car and walked,but did not use a bus
10 students travelled by bus and walked,but did not use a car
3 students used all three forms of travel
18 of the 50 walked
probability that the first one walked = 18/50
now you pick a second out of a population of 49 of whom 17 walked
so
18/50 * 17/49 = 306/2450 = .125
prob first one walked, second did not = (18/50) (32/49) = .235
prob second one walked, first did not is the same
so
.235 * 2 = .47
To find the probability in each case, we need to determine the total number of outcomes and the number of favorable outcomes.
a) To find the probability that both students walked, we need to find the number of outcomes where both students only walked and divide it by the total number of outcomes.
The total number of outcomes is given by choosing 2 students out of 50, which is denoted as "50 choose 2" or written as C(50, 2). This can be calculated as:
C(50, 2) = 50! / (2!(50-2)!) = 50! / (2!48!) = (50 * 49) / (2 * 1) = 1225
The number of favorable outcomes is the number of ways to choose 2 students who only walked. This includes students who travelled by car only, bus only, and those who used all three forms of travel (since walking is common to all three). We need to exclude these cases.
The number of students who used only the car is 12.
The number of students who used only the bus is 7.
The number of students who used all three forms of travel is 3.
So, the number of favorable outcomes is (12 + 7 + 3) = 22.
The probability that both students walked is given by the number of favorable outcomes divided by the total number of outcomes:
P(both students walked) = number of favorable outcomes / total number of outcomes
= 22 / 1225
≈ 0.0179
Therefore, the probability that both students walked is approximately 0.0179 or 1.79%.
b) To find the probability that only one of the students walked, we need to find the number of outcomes where one student walked and the other student did not walk. Similarly, we divide this number by the total number of outcomes.
The number of favorable outcomes can be calculated by finding the number of ways to choose one student who walked only and one student who did not walk. Again, we need to exclude the cases where students travelled by car only, bus only, or used all three forms of travel.
The number of students who used only the car is 12.
The number of students who used only the bus is 7.
The number of students who used all three forms of travel is 3.
So, the number of favorable outcomes is (12 + 7 + 3) * (50 - 1) = 22 * 49 = 1078.
The probability that only one of the students walked is given by the number of favorable outcomes divided by the total number of outcomes:
P(only one student walked) = number of favorable outcomes / total number of outcomes
= 1078 / 1225
≈ 0.8796
Therefore, the probability that only one of the students walked is approximately 0.8796 or 87.96%.