Semi-hard probability question.

[Kaunteya]

I don't know the aswers to the question but it is a Kaplan question that is in all of their books. Question reads:

In the city of x the probabilty of rain is 50%. What is the probabilty that it will rain 3 out of 5 days?

I understand that there are 10 possible outcomes, and that the chance that it will rain on a certain day is 1/2. Out of 5 days then is the probabilty of rain 1/2 x 1/2 x 1/2 which is 1/8 on any given day? This is where I get lost. I can answer coin flipping, seat arrangement questions but this one I can't visualize. Thanks gang.

Kaunteya

[Stuart@KaplanGMAT]

I don't know the aswers to the question but it is a Kaplan question that is in all of their books. Question reads:

In the city of x the probabilty of rain is 50%. What is the probabilty that it will rain 3 out of 5 days?

I understand that there are 10 possible outcomes, and that the chance that it will rain on a certain day is 1/2. Out of 5 days then is the probabilty of rain 1/2 x 1/2 x 1/2 which is 1/8 on any given day? This is where I get lost. I can answer coin flipping, seat arrangement questions but this one I can't visualize. Thanks gang.

Kaunteya

This actually IS a coin flip question - it's exactly the same as:

"If a fair coin is tossed 5 times, what's the probability that it lands on heads exactly 3 times?"

Any time we have a 50% probability binomial distribution we can use the coin flip formula, which is:

nCk/2^n

in which n = # of events and k = # of desired results.

In this case, we have n=5 and k=3, so:

(5!/3!2!)/(2^5) = (5*4/2)/32 = 10/32 = 5/16

For a non-50% distribution, we break the formula up into different parts:

nCk * (prob wanted result)^k * (1-probability wanted result)^(n-k)

For example, if there's a 2/3 chance of rain and we want to know the probability of it raining on exactly 3 out of 5 days:

5C3 * (2/3)^3 * (1/3)^2 = 10 * 8/27 * 1/9 = 80/243

[Kaunteya]

For 50% distribution why do you divide the possible outcome (nCk) by 2^n? If (prob wanted result) is 1/2, why then is it two? For a non-50% distribution you said (prob wanted result)^k * (1-prob wanted result)^(n-k). And in your example of the probabilty of it raining is 2/3, the (prob wanted result)^k = (2/3)^k

I appreciate your help, but I still understand why the 2^n is not 1/2^n, when flipping a coin is 50/50. I can obviously see that 1/2^5 nCk/(1/32) would be 320 which is not one of the answer choices, and does not make sense mathimatically, but rather than just memorizing an equation I need to understand them. Cheers again mate. Too bad you are not here in Montreal, I would have forced you to a pint of Guiness for some hints and tips. Thanks.

Kaunteya

[Stuart@KaplanGMAT]

For 50% distribution why do you divide the possible outcome (nCk) by 2^n? If (prob wanted result) is 1/2, why then is it two? For a non-50% distribution you said (prob wanted result)^k * (1-prob wanted result)^(n-k). And in your example of the probabilty of it raining is 2/3, the (prob wanted result)^k = (2/3)^k

I appreciate your help, but I still understand why the 2^n is not 1/2^n, when flipping a coin is 50/50. I can obviously see that 1/2^5 nCk/(1/32) would be 320 which is not one of the answer choices, and does not make sense mathimatically, but rather than just memorizing an equation I need to understand them. Cheers again mate. Too bad you are not here in Montreal, I would have forced you to a pint of Guiness for some hints and tips. Thanks.

Kaunteya

The formula divides by 2^n, which is the same as multiplying by 1/(2^n).

We could also write the formula as:

nCk * 1/(2^n) = nCk/2^n

[gmat765]

It's a quite simple question. The real question will not so simple.

[tigron]

Interesting answers, but the problem is simpler. There are 6 possible outcomes. It rained on exactly 1 day or 2 days or 3days , 4 days , 5 days or 0 days if it did not rain at all. Since the chances of it raining or not raining on each day are equal, the possible outcomes have equal chance. So the probablity of each is 1/6. So probability of it raining on 3 days = 1/6.

[Ian Stewart]

Interesting answers, but the problem is simpler. There are 6 possible outcomes. It rained on exactly 1 day or 2 days or 3days , 4 days , 5 days or 0 days if it did not rain at all. Since the chances of it raining or not raining on each day are equal, the possible outcomes have equal chance. So the probablity of each is 1/6. So probability of it raining on 3 days = 1/6.

If that logic were correct, then your probability of winning the lottery when you buy one ticket would be 50%, because there are only two outcomes: you win or you lose.

You can only use your approach when each outcome is equally likely. In the question above, it is much more likely that it rains on exactly three days than that it rains five days in a row; the probability it rains five days in a row is not 1/6. Indeed, on each day, there are two possible outcomes, each of which is equally likely, so there are 2*2*2*2*2 = 32 possible outcomes in five days. We can choose three days out of five in 5C3 = 10 ways, so the probability is 10/32 = 5/16.

[giaxou]

Interesting answers, but the problem is simpler. There are 6 possible outcomes. It rained on exactly 1 day or 2 days or 3days , 4 days , 5 days or 0 days if it did not rain at all. Since the chances of it raining or not raining on each day are equal, the possible outcomes have equal chance. So the probablity of each is 1/6. So probability of it raining on 3 days = 1/6.

If that logic were correct, then your probability of winning the lottery when you buy one ticket would be 50%, because there are only two outcomes: you win or you lose.

You can only use your approach when each outcome is equally likely. In the question above, it is much more likely that it rains on exactly three days than that it rains five days in a row; the probability it rains five days in a row is not 1/6. Indeed, on each day, there are two possible outcomes, each of which is equally likely, so there are 2*2*2*2*2 = 32 possible outcomes in five days. We can choose three days out of five in 5C3 = 10 ways, so the probability is 10/32 = 5/16.

This is all clear to me - However, why is 5C3 = 10? Is it not 5!/3! = 20?

[Ian Stewart]

This is all clear to me - However, why is 5C3 = 10? Is it not 5!/3! = 20?

You may be thinking of the permutations formula, rather than the combinations formula, although in that case the number would be larger than 10. There are several different notations used in counting, but most are difficult to type on a standard keyboard, so on internet forums, you'll often see "5C3", which a mathematician would read as "5 choose 3". 5C3 means "the number of sets of 3 you can select from a group of 5, if order is *not* important". In general, we have:

nCr = n!/[r!*(n-r)!]

so

5C3 = 5!/(3!*(5-3)!) = 5!/(3!*2!) = (5*4)/2 = 10

So, from a group of 5 people, we can choose 10 three-person boards of directors; if we want to make a three-topping pizza, and we have five toppings to choose from, there are 10 different pizzas we can make; etc.

You may be thinking of the 'permutations' formula, which is sometimes written using the capital letter P. 5P3 would mean "the number of ways to select three things from a group of five *and* put them in order". You never need a formula for permutations - it is normally much faster to simply count your choices for each position and multiply - but if you were to use a formula, it is:

nPr = n!/(n-r)!

So 5P3 = 5!/(5-3)! = 5!/2! = 5*4*3 = 60

So you can choose a President, Vice-President and Treasurer from five people in 60 ways; if five people run a race, there are 60 different possibilities for the Gold/Silver/Bronze medal winners, etc.

[rnola]

Interesting answers, but the problem is simpler. There are 6 possible outcomes. It rained on exactly 1 day or 2 days or 3days , 4 days , 5 days or 0 days if it did not rain at all. Since the chances of it raining or not raining on each day are equal, the possible outcomes have equal chance. So the probablity of each is 1/6. So probability of it raining on 3 days = 1/6.

If that logic were correct, then your probability of winning the lottery when you buy one ticket would be 50%, because there are only two outcomes: you win or you lose.

You can only use your approach when each outcome is equally likely. In the question above, it is much more likely that it rains on exactly three days than that it rains five days in a row; the probability it rains five days in a row is not 1/6. Indeed, on each day, there are two possible outcomes, each of which is equally likely, so there are 2*2*2*2*2 = 32 possible outcomes in five days. We can choose three days out of five in 5C3 = 10 ways, so the probability is 10/32 = 5/16.

What is the probability of having exactly 3 rainy days in a row for a week if the probability of a rainy day is 20% ?

[knight247]

@rnola
Let R represent a rainy day and N represent a non rainy day

Since we need exactly three consecutive rainy days our sequence of events needs to look like this

RRRNN and all the possible permutations of this. Since the rainy days have to consecutive we'll consider RRR as one unit i.e. X

So, we have
XNN and this can be arranged in 3P3/2! ways as N is repeated twice...3P3/2!=3!/2!=3 Ways


P(R)=20/100=1/5
P(N)=80/100=4/5

Now for RRRNN we have the probability as 1/5*1/5*1/5*4/5*4/5=0.2*0.2*0.2*0.8*0.8=0.00512 times 3 since there are 3 ways in which those rainy days could be arranged=0.01536=1536/100000=[spoiler]48/3125[/spoiler]

[GMATGuruNY]

I posted a solution here. The solution includes a link to a similar but trickier probability problem:

https://www.beatthegmat.com/rain-prob-t88396.html