Can anyone help me with this probability question?

Question

Due to a manufacturing error, 3 cans were accidently filled with diet soda and placed into a 12 pack labeled regular soda. suppose 2 cans are randomly selected from the 12 pack. Determine the probability that exactly one of the cans contains diet soda.
Sorry the question is lengthy. plz reply ASAP.
Thankz

Answer 1

No. of cans with diet soda = 3
No. of cans with regular soda = 9

Let's treat the diet sodas as D and regulars as R
since we are taking out 2 cans at random, they can be...

D D, R R, D R and R D

P (selecting a D as first can) = 3/12
P (selecting an R as second can) = 9/11
because after picking out D as the first can you are left with 11 cans

P (selecting an R as first can) = 9/12
P (selecting a D as second can) = 3/11

P (exactly one can contains diet soda) = 3/12 x 9/11 + 9/12 x 3/11 = 9/44 + 9/44 = 9/22

Hope this helps =)

Answer 2

the probability that you first get a diet soda, then a regular soda is 3/12 * 9/11 = 9 /44

the probability that you first get a regular soda, then a diet soda is
9/12 * 3/11 = 9/44

then the probability that you get exactly one diet soda in random drawing to cans is 9/44 + 9/44 = 9/22

Answer 3

i trust my argument will no longer be nicely received. I say the prospect is 50% enable a ??, enable b ?? and randomly opt for the values for a and b. As already talked about, for a ? 0, P( a < b²) = a million, that's trivial. in person-friendly words somewhat a lot less trivial is the idea that P(a < 0 ) = a million/2 and to that end P( a < b² | a ? 0) = a million and P( a < b² ) ? a million/2 Now evaluate what takes position even as a > 0 For a > 0, even as it really is elementary to exhibit there's a non 0 danger for a finite b, the decrease, the prospect is 0. a < b² is equivalent to keeping 0 < a < b², undergo in ideas we are in person-friendly words searching at a > 0. If this a finite period on an unlimited line. The danger that a is a component of this period is 0. P( a < b² | a > 0) = 0 As such we've a complete danger P( a < b² ) = P( a < b² | a ? 0) * P(a ? 0) + P( a < b² | a > 0) * P(a > 0) = a million * a million/2 + 0 * a million/2 = a million/2 undergo in ideas, that is because of the endless gadgets. no remember what type of period you draw on paper or on a workstation you'll stumble on a finite danger that looks to attitude a million. yet that is because of the finite random variety turbines on the workstation and if we had this question requested with finite values there will be a a answer better than 50%. i do not advise to be condescending, yet please clarify why utilizing the Gaussian to approximate a uniform distribution is a sturdy concept? are not endless numbers interesting. Cantor even as mad operating with them! :)

Answer 4

It doesn't matter which one is chosen first, you'll end up with the same answer.

Say you wanna choose diet first, you have a 1/4 chance. After that, there's a 9/11 chance of getting a regular one. This leads you to a 9/44 chance.

The other way around, you have a 3/4 chance of getting a regular one first and a 3/11 chance of getting a diet one next, also leading to a 9/44 chance.

Edit: Yeah, I completely forgot about the permutation of them. Blast! It's 9/22.

Answer 5

For the first draw the probability of selecting a diet is 3 of 12 and the nondiet is 9 of 12. We draw a can out, and its one of two options.

If its nondiet (9/12) then the probability for the next selection is 3 of 11 for diet and 8 of 11 for nondiet. To select the diet the probability is 3 of 11. The cumulative probability is (9/12)*(3/11).

If its diet (3/12) then the probability for the next selection is 2 of 11 for diet and 9 of 11 for nondiet. To select the nondiet the probability is 9 of 11. The cumulative probability is (3/12)*(9/11).

The probability for both of the processes is (3/12)*(9/11)+(9/12)*(3/11) = 40.9%

Answer 6

just as scuazmooq said, but the answer is actually the sum of those two events, which gives 9/44 + 9/44 = 9/22

Answer 7

Wouldn't those have to be summed, since either condition meets the requirements:

P1(D)*P2(R) + P1(R)*P2(D),

P1(D) = Pr (1st is diet) P2(R) = Pr 2nd is regular
P1(R) = Pr (1st is regular) P2(D) = Pr 2nd is diet

??

Answer 8

I believe it is 0.125. I am not sure, though sorry. I am not good at math. And can u believe i graduated from High school? Dumb me.