70% of the student body of a very large post-scondary institution is female. In a random sample of 12 students, what is the probability that:
a) at most half will be females
b) more than 7 will be females
c) fewer than 3 will be males
I used the Table of Binomial Probabilities:
X P(x) X P(x)
0 .0000 7 .1585
1 .0000 8 .2311
2 .0002 9 .2397
3 .0015 10 .1678
4 .0078 11 .0712
5 .0291 12 .0138
6 .0792
(A) P(at most 1/2 female)=
P(0 or 1 or 2 or 3 or 4 or 5 or 6)
P(x=0)+ P(x=1)+ P(x=2)+ P(x=3)+ P(x=4)+ P(x=5)+ P(x=6)
= .0000+.0000+.0002+.0015+.0078+.0291+.0792= .1178 is this right?
(B) P(more than 7 will be female)=
P(8 or 9 or 10 or 11 or 12)
P(x=8)+ P(x=9)+ P(x=10)+ P(x=11)+ P(x=12)= .2311+.2397+.1678+.0712+.0138= .7236 is this right?
(C) I don't know how to do this one :(
PS I don't want the answer, for C, I just need to understand how to do the formula. Anyone?
Anyone? Really need help on how to do this please :)
OK last time, do I add up P(x) values from the binomial table from 3 through 12 to establish p and then subtract p value from 1 (as I would to get q value) to get answer C)? or would I totally switch the p value from women to men. I would like to keep the integrity of the question as much as possible and keep the p as women, but if I must change then I will.
To solve part (a), you did it correctly by summing up the probabilities of each individual outcome. You correctly calculated P(at most 1/2 female) as 0.1178.
To solve part (b), you also followed the correct approach. You calculated the probabilities of each individual outcome and summed them up. You correctly calculated P(more than 7 females) as 0.7236.
Now let's move on to part (c), where you need to find the probability of having fewer than 3 males in a random sample of 12 students.
To solve this, we can consider the complement event, which is having 3 or more males. We will calculate the probability of this complement event and then subtract it from 1 to get the probability of having fewer than 3 males.
The complement event can occur in four ways: 3 males, 4 males, 5 males, or 6 males.
First, let's find the probability of having exactly 3 males. We can use the binomial probability formula, which states:
P(x) = nCx * p^x * (1-p)^(n-x)
Where:
- n is the number of trials or total sample size (12 students in our case)
- x is the number of successes (males in our case)
- p is the probability of success (probability of selecting a male, which is 30% or 0.3)
- nCx is the binomial coefficient, calculated as n! / (x! * (n-x)!)
For x = 3, we have:
P(x=3) = 12C3 * (0.3)^3 * (1-0.3)^(12-3)
Calculating this will give you the probability of having exactly 3 males. Repeat this process for x = 4, 5, and 6 to find the probabilities of having exactly 4, 5, and 6 males.
Finally, add up the probabilities of these four outcomes to find the probability of the complement event (3 or more males). Subtract this probability from 1 to get the probability of having fewer than 3 males.
P(fewer than 3 males) = 1 - (P(x=3) + P(x=4) + P(x=5) + P(x=6))
Calculate this expression to find the desired probability.