Last visit was: 19 May 2024, 00:24 It is currently 19 May 2024, 00:24

Close

GRE Prep Club Daily Prep

Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GRE score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Close

Request Expert Reply

Confirm Cancel
User avatar
Retired Moderator
Joined: 07 Jun 2014
Posts: 4810
Own Kudos [?]: 10633 [4]
Given Kudos: 0
GRE 1: Q167 V156
WE:Business Development (Energy and Utilities)
Send PM
Retired Moderator
Joined: 07 Jan 2018
Posts: 739
Own Kudos [?]: 1376 [2]
Given Kudos: 93
Send PM
User avatar
Retired Moderator
Joined: 07 Jun 2014
Posts: 4810
Own Kudos [?]: 10633 [1]
Given Kudos: 0
GRE 1: Q167 V156
WE:Business Development (Energy and Utilities)
Send PM
avatar
Intern
Intern
Joined: 26 Sep 2021
Posts: 1
Own Kudos [?]: 1 [1]
Given Kudos: 4
Send PM
A certain city has a chance of rain occurring on any given d [#permalink]
1
sandy wrote:
Explanation

In essence, the question is asking, “What is the probability that one or more days are rainy days?” since any single rainy day would mean the city experiences rain. In this case, employ the 1 – x shortcut, where the probability of rain on one or more days is equal to 1 minus the probability of no rain on any day.

Since the probability of rain is \(\frac{1}{3}\) on any given day, the probability of no rain on any given day is \(1 - \frac{1}{3}=2/3\) .

Therefore, the probability of no rain on three consecutive days is \(\frac{2}{3}\frac{2}{3}\frac{2}{3}=\frac{8}{27}\).

Finally, subtract from 1 to find the probability that it rains on one or more days: P(1 or more days) = 1 – P(no rain) = \(1 - \frac{8}{27}=\frac{19}{27}\).


Why cant we simply multiply 1/3*1/3*1/3 which is probability of rain for any given day?
Retired Moderator
Joined: 02 Dec 2020
Posts: 1833
Own Kudos [?]: 2125 [4]
Given Kudos: 140
GRE 1: Q168 V157

GRE 2: Q167 V161
Send PM
Re: A certain city has a chance of rain occurring on any given d [#permalink]
4
Hey shubhankarsapa

Read the stem very carefully: In any given 3-day period, what is the probability that the city experiences rain?

So when you multiply for rain for each day, you will get the probability that it will rain all 3 days. But what if it rains the third day and not the first two? It will also be considered under the stem. So we need to find the probability of rain on any day not all days.

Hence we calculate by 1 - no rain = raining at least once.

shubhankarsapa wrote:
Why cant we simply multiply 1/3*1/3*1/3 which is probability of rain for any given day?
User avatar
GRE Prep Club Legend
GRE Prep Club Legend
Joined: 07 Jan 2021
Posts: 4503
Own Kudos [?]: 68 [0]
Given Kudos: 0
Send PM
Re: A certain city has a chance of rain occurring on any given d [#permalink]
Hello from the GRE Prep Club BumpBot!

Thanks to another GRE Prep Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
Prep Club for GRE Bot
[#permalink]
Moderators:
Moderator
1085 posts
GRE Instructor
218 posts

Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne