Question 1206772: The Wilson family was one of the first to come to the U.S. They had 7 children. Assuming that the probability of a child being a girl is .5, find the probability that the Wilson family had:
at least 5 girls?
at most 4 girls?
Found 2 solutions by ikleyn, greenestamps:
Answer by ikleyn(52794)   (Show Source):
You can put this solution on YOUR website! .
The Wilson family was one of the first to come to the U.S. They had 7 children.
Assuming that the probability of a child being a girl is .5, find the probability that the Wilson family had:
(a) at least 5 girls?
(b) at most 4 girls?
~~~~~~~~~~~~~~~~~~~~~~~~
This problem is on the Binomial distribution.
The number of trials is 7. The probability of success for each individual trial is 0.5.
(a) In part (a), they want you determine the probability
P(at least 5 girls) = P(5) + P(6) + P(7).
+-----------------------------------------------------------------------+
| You can calculate each of the three terms/addends above separately |
| and then add them. |
| But the shortest way is to calculate P as |
| P = 1 - (P(0) + P(1)+ P(2) + P(3) + P(4)). |
| The sum in parentheses is a cumulative sum, and there are special |
| tools to calculate cumulative sums. |
+-----------------------------------------------------------------------+
To find the cumulative sum in parentheses, use the standard function binomcdf
of a regular calculator TI-83/84
k n p <<<---=== formatting pattern
P' = binomcdf(4, 7, 0.5)
You get P' = 0.7734 (rounded).
Then P = 1 - P' = 1 - 0.7734 = 0.2266. It is the ANSWER to (a).
Notice that P' is the complementary probability to P.
Alternatively, to calculate cumulative sum, you may use the Excel standard function BINOM.DIST.
Another alternative is to use free of charge online calculator
https://stattrek.com/online-calculator/binomial.aspx
This calculator is the best, since it gives an output in all formats simultaneously.
You only should choose the format which you need.
(b) In part (b), they want you determine the probability
P(at most 4 girls) = P(0) + P(1) + P(2) + P(3) + P(4).
+-----------------------------------------------------------------------+
| The sum in parentheses is a cumulative sum, and there are special |
| tools to calculate cumulative sums. |
+-----------------------------------------------------------------------+
To find the cumulative sum in parentheses, use the standard function binomcdf
of a regular calculator TI-83/84
k n p <<<---=== formatting pattern
P = binomcdf(4, 7, 0.5)
You get P = 0.7734 (rounded). It is the ANSWER to (b).
Alternatively, to calculate cumulative sum, you may use the Excel standard function BINOM.DIST.
Another alternative is to use free of charge online calculator
https://stattrek.com/online-calculator/binomial.aspx
Again, this calculator is the best, since it gives an output in all formats simultaneously.
You only should choose the format which you need.
Solved.
Answer by greenestamps(13200)  (Show Source):
You can put this solution on YOUR website!
If you have a problem like this in a situation where you don't have access to a calculator (or where use of a calculator is not allowed), then you can solve the problem using Pascal's Triangle. It can be used in any problem where there are exactly two outcomes with equal probabilities -- like genders of children, or flips of a fair coin.
Note at the beginning that "at least 5 girls" and "at most 4 girls" cover all the possibilities with no overlap. That means the sum of the answers to the two questions must be 1. Alternatively, it means that after you answer one of the questions you can simply subtract that answer from 1 to get the answer to the other.
For this problem, involving the genders of 7 children, we are interested in row 7 of Pascal's Triangle, which is
1 7 21 35 35 21 7 1
Those numbers are the numbers of ways there can be, respectively, 7, 6, 5, 4, 3, 2, 1, or 0 girls.
Then, since the total number of ways of getting 7 children is 2^7=128, the probabilities are
1/128, 7/128, 21/128, 35/128, 35/128, 21/128, 7/128, and 1/128
of getting the respective numbers of girls.
So for this problem, the probability of getting at least 5 girls (i.e., 7 or 6 or 5) is 1/128 + 7/128 + 21/128 = 29/128; and then the probability of getting at most 4 girls is 1-29/128 = 99/128 (or 35/128 + 35/128 + 21/128 + 7/128 + 1/128 = 99/128).
A student who works this kind of problem frequently might have the first several rows of Pascal's Triangle memorized. But the student can generate any row using an easy algorithm.
The numbers in row 7, used for this problem, can be generated like this:
1
1*(7/1) = 7
7*(6/2) = 7*3 = 21
21*(5/3) = 7*5 = 35
35*(4/4) = 35
and after that you know the numbers form a symmetrical pattern; row 7 is
1 7 21 35 35 21 7 1
Here is how to generate the numbers in row 10 using the algorithm:
1
1*(10/1) = 10
10*(9/2) = 9*5 = 45
45*(8/3) = 8*15 = 120
120*(7/4) = 7*30 = 210
210*(6/5) = 6*42 = 252
252*(5/6) = 210
and again use the symmetrical pattern to find that row 10 of Pascal's Triangle is
1 10 45 120 210 252 210 120 45 10 1
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