Events & Promotions
|
|
GMAT 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
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Apr 2808:00 AM PDT
-11:00 AM PDT
Whether you are just beginning your MBA journey or fine-tuning your path to a 700+ score, GMAT Day is designed to give you a competitive edge.
10 Jan 2021, 11:15
Kudos
Bookmarks
B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
85%
(hard)
Question Stats:
46% (02:12) correct
54%
(02:21)
wrong
based on 164
sessions
History
Date
Time
Result
Not Attempted Yet
A box contains three coins, two of them are fair and the third is two-headed. A coin is selected at random and tossed. If heads appears, the coin is tossed again. If tails appears, then another coin is selected from the two remaining coins and tossed. If the same coin is tossed twice, what is the probability that it is the two-headed coin?
A. 2/3
B. 1/2
C. 3/8
D. 1/4
E. 1/8
Are You Up For the Challenge: 700 Level Questions
A. 2/3
B. 1/2
C. 3/8
D. 1/4
E. 1/8
Are You Up For the Challenge: 700 Level Questions
10 Nov 2021, 06:12
Kudos
Bookmarks
The important thing to note here is : the same coin is tossed twice ie, we are not concerned with the outcome of 2nd toss...only 1st should be heads.
1st heads can come in 2 ways ...if 2 fair coins are chosen and both render heads + unfair coin is chosen:
Case 1: Fair coin: 2/3 * 1/2 = 1/3 (2/3 -- prob. of choosing 2 Fair coin, 1/2 - 1st toss heads)
Case 2: unfair coin: 1/3 *1 = 1/3 (1/3 -- prob. of choosing unfair coin, 1 - 1st toss heads)
P = unfair / total
P = 1/3 / (1/3+1/3) = 1/2
B
1st heads can come in 2 ways ...if 2 fair coins are chosen and both render heads + unfair coin is chosen:
Case 1: Fair coin: 2/3 * 1/2 = 1/3 (2/3 -- prob. of choosing 2 Fair coin, 1/2 - 1st toss heads)
Case 2: unfair coin: 1/3 *1 = 1/3 (1/3 -- prob. of choosing unfair coin, 1 - 1st toss heads)
P = unfair / total
P = 1/3 / (1/3+1/3) = 1/2
B
General Discussion
Deepakjhamb
Joined: 29 Mar 2020
Last visit: 15 Sep 2022
Posts: 215
Own Kudos:
Given Kudos: 14
Location: India
Concentration: General Management, Leadership
Schools: HBS '22 IIMA PGPX'22
GPA: 3.96
WE:Business Development (Telecommunications)
10 Jan 2021, 11:46
Kudos
Bookmarks
p (same coin tossed ) * p( 2headed coin /same coin tossed) = p( same coin tossed and 2 head coin)
p(same coin tossed ) = 2/3 *1/2 *1 + 1/3 *1*1 = 1/6+1/3 = 1/2
p(same coin tossed and 2 headed coin) = 1/3 *1 *1 =1/3
p( 2headed coin /same coin tossed) = 1/3/(1/2) = 2/3
so answer is a)2/3
p(same coin tossed ) = 2/3 *1/2 *1 + 1/3 *1*1 = 1/6+1/3 = 1/2
p(same coin tossed and 2 headed coin) = 1/3 *1 *1 =1/3
p( 2headed coin /same coin tossed) = 1/3/(1/2) = 2/3
so answer is a)2/3
