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26 May 2024, 04:55
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Prize 1: 0.5
Prize 2: 0.9
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
75%
(hard)
Question Stats:
63% (02:38) correct
37%
(02:53)
wrong
based on 861
sessions
History
Date
Time
Result
Not Attempted Yet
At a carnival game, a winning player spins a wheel that always lands on either Prize 1 or Prize 2 to determine which of the two prizes he or she wins. The probability that the prize wheel indicates Prize 2 is double the probability that it indicates Prize 1. If a player does not want the prize that the prize wheel first indicates, then he or she may spin the wheel again. In such cases, the player must accept whichever prize the prize wheel indicates on the second spin.
Select for Prize 1 the number nearest to the probability that a winning player who wants Prize 1 will receive Prize 1 after one or two spins of the prize wheel, and select for Prize 2 the number nearest to the probability that a winning player who wants Prize 2 will receive Prize 2 after one or two spins of the prize wheel. Make only two selections, one in each column.
GMAT-Club-Forum-kdeik1nt.png [ 93.18 KiB | Viewed 2088 times ]
Select for Prize 1 the number nearest to the probability that a winning player who wants Prize 1 will receive Prize 1 after one or two spins of the prize wheel, and select for Prize 2 the number nearest to the probability that a winning player who wants Prize 2 will receive Prize 2 after one or two spins of the prize wheel. Make only two selections, one in each column.
Attachment:
GMAT-Club-Forum-kdeik1nt.png [ 93.18 KiB | Viewed 2088 times ]
| Prize 1 | Prize 2 | |
| 0.1 | ||
| 0.3 | ||
| 0.5 | ||
| 0.7 | ||
| 0.9 |
ShowHide Answer
Official Answer
Prize 1: 0.5
Prize 2: 0.9
28 Jun 2024, 12:59
Kudos
Bookmarks
My Approach:
let P(p1) = x, P(p2) = 2x => x+2x = 1 => 3x =1
so x= 1/3.
1. For Prize 1:
Two possible scenarios:
a. Wants p1 - gets in first time - 1/3
b. Gets p2 - but as it wants p1 - spin again and get p1 => 2/3*1/3
Combined => 1/3 + 2/9 = 5/9 => 0.555 = 0.5 (approx.)
2. For Prize 2:
Two possible scenarios:
a. Wants p2 - gets in first time - 2/3
b. Gets p1- but as it wants p2 - spin again and get p1 => 1/3*2/3
Combined => 2/3 + 2/9 = 8/9 => 0.888.. = 0.8 (approx.)
Hope that helps !
Thanks
let P(p1) = x, P(p2) = 2x => x+2x = 1 => 3x =1
so x= 1/3.
1. For Prize 1:
Two possible scenarios:
a. Wants p1 - gets in first time - 1/3
b. Gets p2 - but as it wants p1 - spin again and get p1 => 2/3*1/3
Combined => 1/3 + 2/9 = 5/9 => 0.555 = 0.5 (approx.)
2. For Prize 2:
Two possible scenarios:
a. Wants p2 - gets in first time - 2/3
b. Gets p1- but as it wants p2 - spin again and get p1 => 1/3*2/3
Combined => 2/3 + 2/9 = 8/9 => 0.888.. = 0.8 (approx.)
Hope that helps !
Thanks
General Discussion
26 May 2024, 11:03
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sreehere
At a carnival game, a winning player spins a wheel that always lands on either Prize 1 or Prize 2 to determine which of the two prizes he or she wins. The probability that the prize wheel indicates Prize 2 is double the probability that it indicates Prize 1. If a player does not want the prize that the prize wheel first indicates, then he or she may spin the wheel again. In such cases, the player must accept whichever prize the prize wheel indicates on the second spin.
Select for Prize 1 the number nearest to the probability that a winning player who wants Prize 1 will receive Prize 1 after one or two spins of the prize wheel, and select for Prize 2 the number nearest to the probability that a winning player who wants Prize 2 will receive Prize 2 after one or two spins of the prize wheel. Make only two selections, one in each column.
Posted from my mobile device
As the Source and OA are correct now, the solution would be Select for Prize 1 the number nearest to the probability that a winning player who wants Prize 1 will receive Prize 1 after one or two spins of the prize wheel, and select for Prize 2 the number nearest to the probability that a winning player who wants Prize 2 will receive Prize 2 after one or two spins of the prize wheel. Make only two selections, one in each column.
Posted from my mobile device
The prize wheel can be taken to be consisting of three parts, out of which prize 1 one occupies one part and prize 2 occupies the remaining two parts.
(1) Prize 1 after one or two spins-
First way: Total P - no prize 1 after two spins =\( 1 - (1-\frac{1}{3})(1-\frac{1}{3})=1-\frac{4}{9}=\frac{5}{9} \)or 55.55% closest to 0.5 in options.
Second way: P in first spin + P in second spin =\(\frac{ 1}{3} + (\frac{2}{3}*\frac{1}{3}) = \frac{5}{9}\)
(2) Prize 2 after one or two spins-
First way: Total P - no prize 2 after two spins =\( 1 - (1-\frac{2}{3})(1-\frac{2}{3})=1-\frac{1}{9}=\frac{8}{9} \) or 88.88% closest to 0.9 in options.
Second way: P in first spin + P in second spin = \(\frac{2}{3 }+ (\frac{1}{3}*\frac{2}{3}) = \frac{8}{9}\)
bhattarbharat
