Last visit was: 05 Nov 2024, 08:25 It is currently 05 Nov 2024, 08:25

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
avatar
Manager
Manager
Joined: 22 Jul 2018
Posts: 80
Own Kudos [?]: 104 [3]
Given Kudos: 0
Send PM
avatar
Manager
Manager
Joined: 22 Jul 2018
Posts: 80
Own Kudos [?]: 104 [0]
Given Kudos: 0
Send PM
avatar
Intern
Intern
Joined: 21 Dec 2018
Posts: 20
Own Kudos [?]: 8 [0]
Given Kudos: 0
Send PM
Moderator
Moderator
Joined: 02 Jan 2020
Status:GRE Quant Tutor
Posts: 1111
Own Kudos [?]: 960 [1]
Given Kudos: 9
Location: India
Concentration: General Management
Schools: XLRI Jamshedpur, India - Class of 2014
GMAT 1: 700 Q51 V31
GPA: 2.8
WE:Engineering (Computer Software)
Send PM
A fair die with sides numbered 1, 2, 3, 4, 5, and 6 is to be [#permalink]
1
Given that A fair die with sides numbered 1, 2, 3, 4, 5, and 6 is to be rolled until a number greater than 4 first appears and We need to find What is the probability that a number greater than 4 will first appear on the third or fourth roll?

At each stage The Probability of Getting a number Greater than 4 = \(\frac{2}{6}\) (As there are two numbers greater than 4 i.e. 5 and 6 out of 1 to 6) = \(\frac{1}{3}\)

Similarly, Probability of Getting a number less than or equal to 4 = \(\frac{4}{6}\) = \(\frac{2}{3}\)

Probability that a number greater than 4 will first appear on the third or fourth roll = P(number greater than 4 will first appear on the third roll) + P(number greater than 4 will first appear on the fourth roll)

P(number greater than 4 will first appear on the third roll) = P(First Roll will give ≤ 4) * P(Second Roll will give ≤ 4) * P(Third Roll will give > 4) = \(\frac{2}{3}\) * \(\frac{2}{3}\) * \(\frac{1}{3}\) = \(\frac{4}{27}\)

P(number greater than 4 will first appear on the fourth roll) = P(First Roll will give ≤ 4) * P(Second Roll will give ≤ 4) * P(Third Roll will give ≤ 4) * P(Fourth Roll will give > 4) = \(\frac{2}{3}\) * \(\frac{2}{3}\) * \(\frac{2}{3}\) *\(\frac{1}{3}\) = \(\frac{8}{81}\)

Probability that a number greater than 4 will first appear on the third or fourth roll = P(number greater than 4 will first appear on the third roll) + P(number greater than 4 will first appear on the fourth roll) = \(\frac{4}{27}\) + \(\frac{8}{81}\) = \(\frac{(12 + 8)}{81}\) = \(\frac{20}{81}\)

So, Answer will be E
Hope it helps!

Watch the following video to learn How to Solve Dice Rolling Probability Problems

Prep Club for GRE Bot
A fair die with sides numbered 1, 2, 3, 4, 5, and 6 is to be [#permalink]
Moderators:
GRE Instructor
77 posts
GRE Forum Moderator
37 posts
Moderator
1111 posts
GRE Instructor
228 posts

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