Probability with Combinations

[katty]

Hi,
In one of the Princeton Review online tests:

There is a 90% chance that a registered voter in Burghtown voted in the last election. If five registered voters are chosen at random, what is the approximate likelihood that exactly four of them voted in the last election?

Okay, so I know that first, you have to find the total number of possible outcomes. Except, I don't know how to find that! I know that if it was about tossing coins, the probability would be 50% so for 5 coins, it would be 32 outcomes. But since this is a 90% chance, I'm not sure how to approach it. Please help!

[earth@work]

Hi,
Let E be the event person voted
P(E)=0.9
P(not E)=1-0.9 =0.1
now, we have to find prob of 4 people voted out of 5 selected
Let A,B,C,D,E be 5 people
P(ABCD voted &E did not vote)=0.9*0.9*0.9*0.9*0.1=(0.9)^4*0.1
also, this can happen in 5 ways (i.e ABCE voted and not D etc )
P(exactly 4 voted) =(0.9)^4*0.1 *5 = 0.328 approx
Do let me know correct answer.
cheers!

[katty]

Yes that is the correct answer. But I'm just trying to wrap my head around how I could use combinations for this kind of question. I had another similar question before which was:

The probability that it will rain in town A is 50%. What is the probability that it will rain exactly 3 days out of a 5-day period?

Here, I calculated that there are two possible outcomes for each day - rain or no rain - and then multiplied it by 5 to get 32 outcomes. Then using combinations, 5 choose 3, I got 10 possible ways it could rain for exactly 3 days in a 5-day period. Thus, 10/32 was the answer.

How can I do the same thing for the above voting question? I feel like I can't seem to get probabilities and its really frustrating me...

[katty]

Sorry, never mind. I just worked out how to get the same answer using your method and i think it's a lot easier. Thanks!!!!

[earth@work]

Yes that is the correct answer. But I'm just trying to wrap my head around how I could use combinations for this kind of question. I had another similar question before which was:

The probability that it will rain in town A is 50%. What is the probability that it will rain exactly 3 days out of a 5-day period?

Here, I calculated that there are two possible outcomes for each day - rain or no rain - and then multiplied it by 5 to get 32 outcomes. Then using combinations, 5 choose 3, I got 10 possible ways it could rain for exactly 3 days in a 5-day period. Thus, 10/32 was the answer.

How can I do the same thing for the above voting question? I feel like I can't seem to get probabilities and its really frustrating me...

Hi,
i doubt we can treat these as coin flipping prob questions, coz these events r not equally likely events. The above question worked coz the Probability of raining was luckily 50%(i.e rain was equally likely here) , so your taking 2^5 as total outcome worked. If the it was not 50% the solution above wud not have worked.
Here , we again need to take P(rain) =0.5, P(not rain)=0.5
P(rain exactly 3 days)=(0.5)^3 *(0.5)^2 *5C3 = (0.5)^5 * 10 =0.312 which is same as 10/32
passing u link for probabilities which might help u! i received this from someone here at beatthegmat ... was helpful so passing it on..hope it helps
https://people.richland.edu/james/lecture/m170/

[drpawan]

Hi,
In one of the Princeton Review online tests:

There is a 90% chance that a registered voter in Burghtown voted in the last election. If five registered voters are chosen at random, what is the approximate likelihood that exactly four of them voted in the last election?

Okay, so I know that first, you have to find the total number of possible outcomes. Except, I don't know how to find that! I know that if it was about tossing coins, the probability would be 50% so for 5 coins, it would be 32 outcomes. But since this is a 90% chance, I'm not sure how to approach it. Please help!

p= voting of a registered voter in last election = 90/100 = 9/10
q = negation of p = 1/10
As per binomial probability function,
The probability of getting exactly x success in n trials, with the probability of success on a single trial being p is:
P(X=x) = nCx * p^x * q^(n-x)
P(x=4) = 5C4 * (9/10)^4 * (1/10)^(5-4)
= 5*(9/10)^4*1/10
= 1/2*(9/10)^4
= 3270.5/10000
= 0.327(app)

[stop@800]

Yes that is the correct answer. But I'm just trying to wrap my head around how I could use combinations for this kind of question. I had another similar question before which was:

The probability that it will rain in town A is 50%. What is the probability that it will rain exactly 3 days out of a 5-day period?

Here, I calculated that there are two possible outcomes for each day - rain or no rain - and then multiplied it by 5 to get 32 outcomes. Then using combinations, 5 choose 3, I got 10 possible ways it could rain for exactly 3 days in a 5-day period. Thus, 10/32 was the answer.

How can I do the same thing for the above voting question? I feel like I can't seem to get probabilities and its really frustrating me...

Hi,
i doubt we can treat these as coin flipping prob questions, coz these events r not equally likely events. The above question worked coz the Probability of raining was luckily 50%(i.e rain was equally likely here) , so your taking 2^5 as total outcome worked. If the it was not 50% the solution above wud not have worked.
Here , we again need to take P(rain) =0.5, P(not rain)=0.5
P(rain exactly 3 days)=(0.5)^3 *(0.5)^2 *5C3 = (0.5)^5 * 10 =0.312 which is same as 10/32
passing u link for probabilities which might help u! i received this from someone here at beatthegmat ... was helpful so passing it on..hope it helps
https://people.richland.edu/james/lecture/m170/

We can treat it like a coin flipping question but we need to take the coin as biased coin.