a group of 6 men and 8 women and 5 people are chosen at random,
find probability that:
a) There are exactly 3 men.
b)There are at most 2 woman.
c) There are more men than woman.
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There are three ways to do binomial problems:
1. By a probability table
2. By using a TI-83 or 84 calculator
3. By formula
You didn't specify which way, so I'll do it the easiest way,
with a TI-83 or 84 calculator. If your teacher doesn't allow the
use of those calclators, then post again, stating which of the other
two methods you are to use.
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When using the calculator, the trick is knowing whether to use
binompdf or binomcdf. The "p" is for a particular value of x,
and the "c" is for the cumulative value from 0 up to the value of x.
n = number of trials = 5
p = 6/14 = 3/7 is the probability of selecting 1 man in 1 trial
p = 8/14 = 4/7 is the probability of selecting 1 woman in 1 trial
x = number of men or the number of women depending on the problem
a) There are exactly 3 men.
n = 5, p = 3/7, x = 3
This is a particular value, 3 for x, so we use binompdf:
Press 2nd, then VARS, then scroll down to binompdf( and press ENTER
On the later models TI-84 you are asked for the trials, which is 5,
p: which is 3/7, and the x value which is 3, then you scroll to paste and
press ENTER, you'll see:
binompdf(5,3/7,3
)
On older models and the TI-83 you type "3/7,3)" so that you have the above.
Then you press ENTER and read
.2570357589
b) There are at most 2 women.
n = 5, p = 4/7, x = 2
This is a cumulative value, for 0, 1, or 2 women, so we use binomcdf
We have a maximum value of 2 for x:
Press 2nd, then VARS, then scroll down to binomcdf( and press ENTER
On the later models TI-84 you are asked for the trials, which is 5,
p: which is 4/7, and the x value which is 2, then you scroll to paste and
press ENTER, you'll see:
binomcdf(5,4/7,2
)
On older models and the TI-83 you type "4/7,2)" so that you have the above.
Then you press ENTER and read
.3678824299
c) There are more men than woman.
That means there are 3 men and 2 women, 4 men and 1 woman,
or 5 men and 0 women:
This is the same as problem b, which is also 0, 1, or 2 women. So
the answer is the same.
Edwin
