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# If the probability of rain on any given day in City X is 25%, what is

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If the probability of rain on any given day in City X is 25%, what is  [#permalink]

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29 Sep 2011, 23:17
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If the probability of rain on any given day in City X is 25%, what is probability that it rains on exactly 3 days in a 4 day period.

A. 1/4
B. 1/32
C. 3/64
D. 4/32
E. 3/4
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Posts: 335
Re: If the probability of rain on any given day in City X is 25%, what is  [#permalink]

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30 Sep 2011, 02:39
2
In this question we have to understand its made of two steps:
1st step - what is the probability for 3 days with rain and 1 without
so we have 1/4*1/4*1/4*(1-1/4) = 3/256

The second step however is to understand that we have more than one way to arrange 3 days of rain out of 4. We actually have 4 possible ways to arrange it.
3C4 = 4
or we can just see it by hand bc the number is very small
RRRN
RRNR
RNRR
NRRR

so we need to multiply 4 in the original probability so we will have 4*3/256 ----> 3/64

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Re: If the probability of rain on any given day in City X is 25%, what is  [#permalink]

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30 Sep 2011, 18:53
1
this is a binomial distribution problem.

P(rain) = 1/4

getting rain on 3 days out of 4 = 4c3*((1/4)^3)*(3/4) = 3/64

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Re: If the probability of rain on any given day in City X is 25%, what is  [#permalink]

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02 Oct 2011, 11:02
1
3 days can be choosen out of 4 in 4C3 ways.
Probability of raining 3 days out of 4 = 4C3 * (1/4)^3 * (3/4) = 3 / 64
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Re: If the probability of rain on any given day in City X is 25%, what is  [#permalink]

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11 Jan 2017, 07:45
prashantbacchewar wrote:
If the probability of rain on any given day in City X is 25%, what is probability that it rains on exactly 3 days in a 4 day period.

A. 1/4
B. 1/32
C. 3/64
D. 4/32
E. 3/4

$$P(r) = \frac{1}{4}$$ & $$P(r') = \frac{3}{4}$$

So, Required Probability is $$= (\frac{1}{4})^3*\frac{3}{4}$$

Or, Required Probability is $$= \frac{3}{64}$$

Hence, answer will be $$(C) \frac{3}{64}$$
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Re: If the probability of rain on any given day in City X is 25%, what is  [#permalink]

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11 Jan 2017, 10:39
Top Contributor
2
Let me first give you a brief summary of how to tackle probability questions of this kind and then let me work out the question as an example.

By definition, Probability = (Number of favorable outcomes)/(total number of outcomes)

Consider that we have a bag containing 2 blue, 3 green and 4 yellow balls and have to select one ball from the bag then the total outcomes will be 2 + 3 + 4 = 9. If we need to find the probability of selecting one blue ball P(B) from the bag, then P(B) = 2/9, the favorable outcomes here being the 2 blue balls and the total outcomes being all the 9 balls. Now this procedure holds good when we select one ball at a time (selecting one object at a time is referred to as a single event), but things change when we start selecting more than one ball (selecting more than one object at a time is referred to as a complex event).

Lets consider that we now select two balls at a time (complex event). Now this selection can be done in any one of three ways
1. With Replacement : Involves picking the first ball, replacing it, and then picking the second
2. Without Replacement : Involves picking the first ball, NOT replacing it, and then picking the second
3. Simultaneous : Involves picking both balls at once

For the above mentioned 3 cases the universal formula that you can use is

P(complex) = P(product of individual events) * arrangement

Now considering the example of a bag with 2 blue, 3 green and 4 yellow balls, if we need to find the probability of getting the first ball blue and the second ball green when the balls are selected one by one with replacement then

P(BG) = P(B) * P(G) * arrangement ----> P(BG) = 2/9 * 3/9 * 1 = 2/27.

Here the arrangement is considered as 1 since the order here is specified i.e the first ball needs to be blue and the second green.

If the same question were rephrased as find the probability of getting one blue and one green when the balls are selected one by one with replacement then

P(BG) = P(B) * P(G) * arrangement ----> P(BG) = 2/9 * 3/9 * 2!

The 2! here represents the arrangement of the word BG. If we have a word MISSISSIPPI then the arrangements will be 11!/(4!4!2!) where 11! represents the number of alphabets, 4! represents the number of I's, 4! represents the number of S's and 2! represents the number of P's.

The arrangement here takes care of the two possible cases of getting a blue and a green i.e. P(BG) and P(GB).

If the same question were rephrased as find the probability of getting one blue and one green when the balls are selected one by one without replacement then

P(BG) = P(B) * P(G) * arrangement----> P(BG) = 2/9 * 3/8 * 2!

When we deal with a simultaneous selection, the probability way of solving is to treat a simultaneous selection as one by one without replacement and then use the above illustrated procedure.

Now coming to the question that you have posted

If the probability of rain on any given day in City X is 25%, what is probability that it rains on exactly 3 days in a 4 day period.

Let the probability that it rains be P(R) = 1/4 and the probability that it does not rain be P(N) = 3/4

Now we need the probability that it rains on exactly 3 days in a 4 day period. This can be written as complex event P(RRRN)

P(RRRN) = 1/4 * 1/4 * 1/4 * 3/4 * 4!/3!

Simplifying we get 3/64

Hope this helps!

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Re: If the probability of rain on any given day in City X is 25%, what is  [#permalink]

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17 Feb 2017, 06:40
Abhishek009 wrote:
prashantbacchewar wrote:
If the probability of rain on any given day in City X is 25%, what is probability that it rains on exactly 3 days in a 4 day period.

A. 1/4
B. 1/32
C. 3/64
D. 4/32
E. 3/4

$$P(r) = \frac{1}{4}$$ & $$P(r') = \frac{3}{4}$$

So, Required Probability is $$= (\frac{1}{4})^3*\frac{3}{4}$$

Or, Required Probability is $$= \frac{3}{64}$$

Hence, answer will be $$(C) \frac{3}{64}$$

Required Probability is $$= (\frac{1}{4})^3*\frac{3}{4}$$ = 3/256?

You are missing a required combination value. 4c3 * (\frac{1}{4})^3*\frac{3}{4}[/m] = 3/64
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Re: If the probability of rain on any given day in City X is 25%, what is  [#permalink]

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12 Nov 2018, 07:16
prashantbacchewar wrote:
If the probability of rain on any given day in City X is 25%, what is probability that it rains on exactly 3 days in a 4 day period.

A. 1/4
B. 1/32
C. 3/64
D. 4/32
E. 3/4

We need to determine the probability of having 3 rainy days within a 4-day period. We are given that the probability of a rainy day is 1/4, and thus the probability of a non-rainy day is 3/4.

We can assume the first 3 days are rainy (R) and the last day is not rainy (N). Thus:

P(R-R-R-N) = 1/4 x 1/4 x 1/4 x 3/4 = 3/256

However, we need to determine in how many ways it can rain 3 out of 4 days. That number will be equivalent to how many ways we can arrange the letters R-R-R-N.

We use the indistinguishable permutations formula to determine the number of ways to arrange R-R-R-N: 4!/3! = 4 ways

Each of these 4 ways has the same probability of occurring. Thus, the total probability is:

4(3/256) = 12/526 = 3/64

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Re: If the probability of rain on any given day in City X is 25%, what is &nbs [#permalink] 12 Nov 2018, 07:16
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