Why does P and C yield two different answers here

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hendle

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The Greenwood High School Speech and Debate Club hosted its annual car wash fundraiser. Each club member brought a bottle of car wash soap, so there were 11 total bottles. 5 of the bottles contained blue soap.
If a club member randomly selects 7 bottles to pour into the first soap bucket, what is the probability that exactly 2 of the chosen bottles contain blue soap?
In this questions, the answer of using P and using C is different, but i figure it should be the same since both the favourable event size and total event size increases. Besides, why sometimes P and C yield the same answer but sometimes it doesn't?
 
I don't know what you mean by "P" and "C" because there is no mention of either in the problem and you don't define them! Nor, since you have not given an answer nor shown how you have tried to solve it, do I understand why you say "sometimes P and C yield the same answer but sometimes it doesn't". Your whole question makes no sense!

Here is how I would do the problem stated: There are a total of 11 bottles, 5 containing blue soap. The probability the first bottle chosen is blue soap is 5/11. Given that, there are 10 bottles, 4 containing blue soap. The probability the second bottle chosen is blue soap is 4/10= 2/5. Then there are 9 bottles left, 3 containing blue soap. The probability the third bottle does NOT contain blue soap is 6/9= 2/3. Similarly the probability the fourth, fifth, sixth, and seventh bottles do NOT contain blue soap are 5/8, 4/7, 3/6= 1/2, and 2/5. So the probability the seven bottles chosen are two blue soap and then 5 not blue soap, in that order, is (5/11)(4/10)(6/9)(5/8)(4/7)(3/6)(2/5)= (5/11)(2/5)(2/3)(5/8)(4/7)(1/2)(2/5)= (2/11)(1/3)(5/2)(1/7)(2/5)= (1/11)(1/3)(1/7)(2)= 2/231. Doing the same calculation, for the various possible orders, you should see that the different orders (first and fifth bottles blue soap, third and sixth bottles blue soap, etc.) are exactly the same- you have the same integers in numerator and denominator, just in different orders) have exactly the same probability- so we just need to multiply 2/21 by the number of different orders. And you should have learned ("combinatorics") that the number of different ways to order 11 things, 2 the same, the other 9 different from the first two, is \(\displaystyle \frac{11!}{2!9!}= \frac{11(10)}{2}= 11(5)= 55\).

The probability that, of the 7 bottles chosen, exactly two are blue soap is 55(2/231)= 110/231.
 
HallsofIvy said:
I don't know what you mean by "P" and "C" because there is no mention of either in the problem and you don't define them! Nor, since you have not given an answer nor shown how you have tried to solve it, do I understand why you say "sometimes P and C yield the same answer but sometimes it doesn't". Your whole question makes no sense!

Here is how I would do the problem stated: There are a total of 11 bottles, 5 containing blue soap. The probability the first bottle chosen is blue soap is 5/11. Given that, there are 10 bottles, 4 containing blue soap. The probability the second bottle chosen is blue soap is 4/10= 2/5. Then there are 9 bottles left, 3 containing blue soap. The probability the third bottle does NOT contain blue soap is 6/9= 2/3. Similarly the probability the fourth, fifth, sixth, and seventh bottles do NOT contain blue soap are 5/8, 4/7, 3/6= 1/2, and 2/5. So the probability the seven bottles chosen are two blue soap and then 5 not blue soap, in that order, is (5/11)(4/10)(6/9)(5/8)(4/7)(3/6)(2/5)= (5/11)(2/5)(2/3)(5/8)(4/7)(1/2)(2/5)= (2/11)(1/3)(5/2)(1/7)(2/5)= (1/11)(1/3)(1/7)(2)= 2/231. Doing the same calculation, for the various possible orders, you should see that the different orders (first and fifth bottles blue soap, third and sixth bottles blue soap, etc.) are exactly the same- you have the same integers in numerator and denominator, just in different orders) have exactly the same probability- so we just need to multiply 2/21 by the number of different orders. And you should have learned ("combinatorics") that the number of different ways to order 11 things, 2 the same, the other 9 different from the first two, is \(\displaystyle \frac{11!}{2!9!}= \frac{11(10)}{2}= 11(5)= 55\).

The probability that, of the 7 bottles chosen, exactly two are blue soap is 55(2/231)= 110/231.
Click to expand...
oh P and C as in permutation and combination
 
hendle said:
The Greenwood High School Speech and Debate Club hosted its annual car wash fundraiser. Each club member brought a bottle of car wash soap, so there were 11 total bottles. 5 of the bottles contained blue soap. If a club member randomly selects 7 bottles to pour into the first soap bucket, what is the probability that exactly 2 of the chosen bottles contain blue soap?
In this questions, the answer of using P and using C is different
Click to expand...
I assume that P stands for permutation and C for combination. However, the statement clearly implies that we should use combinations.
\(\dfrac{C_2^5\cdot C_5^6}{C_7^{11}}=\dfrac{2}{11}\). SEE HERE
Suppose the question were, the club has five boys and six girls. The sponsor randomly selects seven to go to a debate. What is the probability of her choosing exactly two boys? Is that not the same exact question?
 
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