SOLUTION: Out of 100 students, (49 girls, 51 boys) if three children are selected what is the probability that at least one is a girl?
I am missing a key point to this equation because I ke
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Question 530501: Out of 100 students, (49 girls, 51 boys) if three children are selected what is the probability that at least one is a girl?
I am missing a key point to this equation because I keep getting 49/97 which I believe is wrong
Thanks for any help
Found 2 solutions by stanbon, oberobic:
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
Out of 100 students, (49 girls, 51 boys) if three children are selected what is the probability that at least one is a girl?
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P(at least one girl) = 1 - P(3 boys)
---
= 1 - 51C3/100C3
---
= 1 - 0.1288
---
= 0.8712
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Cheers,
Stan H.
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Answer by oberobic(2304) (Show Source): You can put this solution on YOUR website!
The probability of picking a girl on the first pick is: 49/100.
.
The probability of picking a girl on the second pick depends on whether a girl was picked first.
If a girl was picked initially, then the probability of picking a girl on the second pick is : 48/99.
If a boy was picked initially, then the probability of picking a girl on the second pick is: 49/99.
.
The probability of picking a girl first and second is: 49/100 * 48/99.
.
And so forth.
.
You need to determine the conditional probabilities to find the chance of picking at least 1 girl.
.
| 1st Pick
| 2nd Pick
| 3rd Pick | Probability
| Description
|
| P(girl)=49/100
| P(girl)=48/99
| P(girl)=47/98
| P=0.113939394
| (A) 3 girls
|
| "
| "
| P(boy)=51/98
| P=0.123636364
| (B) 2 girls, 1 boy
|
| "
| P(boy)=51/99
| P(girl)=48/98
| P=0.123636364
| (C) 2 girls, 1 boy
|
| "
| "
| P(boy)=50/98
| P=0.128787879
| (D) 1 girl, 2 boys
|
| P(boy)=51/100
| P(girl)=49/99
| P(girl)=48/98
| P=0.123636364
| (E) 2 girls, 1 boy
|
| "
| "
| P(boy)=50/98
| P=0.128787879
| (F) 1 girl, 2 boys |
| "
| P(boy)=50/99
| P(girl)=49/98
| P=0.128787879
| (G) 1 girl, 2 boys
|
| "
| "
| P(boy)=49/98
| P=0.128787879
| (H) 3 boys
|
The total probability = 1.00, as you can see.
.
P(at least 1 girl) = A+B+C+D+E+F+G = 0.871212121
.
OR
.
You could see the problem in terms of the chance of picking 3 boys. If you do not pick 3 boys, then you have to pick at least 1 girl.
.
P(picking 3 boys) = 51/100 * 50/99 * 49/98 = 0.128787879
.
P(picking at least 1 girl) = 1.0 - 0.128787879 = 0.871212121
.
Same answer. Two ways to get there.
.
Done.
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