Aptitude Discussion

Q. |
Marie is getting married tomorrow, at an outdoor ceremony in the desert. In recent years, it has rained only 5 days each year. Unfortunately, the weatherman has predicted rain for tomorrow. When it actually rains, the weatherman correctly forecasts rain $90%$ of the time. When it doesn't rain, he incorrectly forecasts rain $10%$ of the time. What is the probability that it will rain on the day of Marie's wedding? |

✖ A. |
0.567 |

✔ B. |
0.111 |

✖ C. |
0.332 |

✖ D. |
0.732 |

**Solution:**

Option(**B**) is correct

The sample space is defined by two mutually-exclusive events - it rains or it does not rain. Additionally, a third event occurs when the weatherman predicts rain. Notation for these events appears below.

Event $A_1$. It rains on Marie's wedding.

Event $A_2$. It does not rain on Marie's wedding.

Event $B$. The weatherman predicts rain.

In terms of probabilities, we know the following:

\(P(A_1)=\dfrac{5}{365}=0.0136985\) [It rains 5 days out of the year.]

\(P(A_2)=\dfrac{360}{365}=0.9863013\) [It does not rain 360 days out of the year.]

\(P(B|A_1)=0.9\) [When it rains, the weatherman predicts rain $90\%$ of the time.]

\(P(B|A_2)=0.1\) [When it does not rain, the weatherman predicts rain $10\%$ of the time.]

We want to know $P(A_1|B)$, the probability it will rain on the day of Marie's wedding, given a forecast for rain by the weatherman. The answer can be determined from Bayes' theorem, as shown below.

\(P(A_1|B)=\dfrac{P(A_1)P(B|A_1)}{P(A_1)P(B|A_1)+P(A_2)P(B|A_2)}\)

\(P(A_1|B)=\dfrac{0.014\times 0.9}{0.014\times 0.9+0.986\times 0.1}\)

\(P(A_1|B)=0.111\)

Note the somewhat unintuitive result. Even when the weatherman predicts rain, it only rains only about $11\%$ of the time. Despite the weatherman's gloomy prediction, there is a good chance that Marie will not get rained on at her wedding.

**Poonam Pipaliya**

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Hey Poonam,

$P\left(\dfrac{B}{A_1}\right)$ is conditional probability. It is read as Probability of occuring event $B$ when event $A_1$ has already occured.

Also you may want to check this out:

www.lofoya.com/Aptitude-Questions-and-Answers/Probability/intro

See toward the bottom for Baye's Theorem.

**Rashid**

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The answer should be $= \left(\dfrac{9}{10}\right) \times \left(\dfrac{5}{365}\right) = 0.012328$

The answer you gave would have been correct if the question was - "it rained on marie's wedding, what is the propability that weatherman had correctly forecasted about weather on marie's wedding"

No Rashid,

That would have been $P(B|A_1)=0.9$.

I believe the solution is correct. Bayes theorem is mostly used to find out reverse probability but here solution seems to be okay to me.

Yeah you are right derek, Baye's theorem is used to find out reverse probability but in this question...First event is forecasting by weatherman and the probability is asked for the later event that is the rain on wedding. So it is not he case of reverse probability.

Hey Rashid I guess we are answer calculates the probability of rain on marie's wedding day given weatherman has announce forecast about this.

Isn't it what we need to solve?

Yeah you are right derek and to solve this we do not need baye's theorem. Baye's theorem is used when probability is asked for happening of "earlier" event given "later" event.

Earlier event - about forecast

Later event - to check whether it rains or not.

But in this question "earlier" event is given and probability is asked for "later" event. So no baye's theorem. Is it clear now.

Let me explain this to you with the help of an e.g.

Box 1- 5 red and 7 black balls

Box 2- 4 red and 3 black balls

Q1. A ball is selected from a box. What is the probability that the ball is red given that it is selected from box 1.

Solution 1.

You can see clearly

events can happen only in the following manner

First event - selection of box

Second event - checking the color

irrespective of what probability has been asked.

No Bayes theorem is req. in this question as probability is asked for later (second) event given earlier (first) event.

Q2 A ball is selected from a box. What is the probability that the ball is selected from box 1 given that the ball is red in color.

As i said in the solution of question 1. event will still happen in the same manner irrespective of what probability has been asked.

In this question the probability is asked for earlier (first) event given later (second) event. So Bayes theorem is req.

**Parichat**

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There's some mistakes on $P(B) =\dfac{360}{365}$ not equals what it's written in 1st quesion.

Thank you Parichat, typo has been corrected.

**Human**

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i like this question so much

**Ak**

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X) When it actually rains -> [X-A] the weatherman [X-B] correctly forecasts rain 90% of the time.

Y) When it doesn't rain -> [Y-A] he [Y-B]incorrectly forecasts rain 10% of the time

You can see clear distinction in X-B and Y-B and yet you choose to simply calculate the given %.

I mean if only Y-B would have been "correctly" then you could have taken the event of his rain forecast to be the given %.

This is what you have mentioned: "P(B|A2)=0.1 [When it does not rain, the [Z-B] "weatherman predicts rain" 10% of the time.]"

Z-B does actually sound like in this step you have assumed it to be "correctly predicts" but that is not what Y-B is.

Could you please explain the same?

**Pradeep**

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i like this problem very much.so please upload the more questions.

**Prasad**

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Great

I love this keep it up and upload more sums.

i can't understand from the formula $P\left(\dfrac{B}{A_1}\right)$,and what is it's formula and what is bayes formula...

please elaborate it..