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Dropdown 1: 1/2
Dropdown 2: 7/8
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
55%
(hard)
Question Stats:
18% (01:33) correct
82%
(01:39)
wrong
based on 707
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A player rolls a fair six-sided die with faces numbered 1 through 6. If the result is even, the player wins a prize and does not roll again. If the result is odd, the player rolls again. The player may roll the die up to three times in total, but stops immediately upon getting an even number. The diagram shows the outcomes and the corresponding prize amounts.
From each drop-down menu, select the option that creates the most accurate statement based on the information provided.
The probability that the player wins $150 is and the probability that the player wins at least $50 is .
ShowHide Answer
Official Answer
Dropdown 1: 1/2
Dropdown 2: 7/8
12 May 2025, 03:55
Kudos
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Official Solution:
A player rolls a fair six-sided die with faces numbered 1 through 6. If the result is even, the player wins a prize and does not roll again. If the result is odd, the player rolls again. The player may roll the die up to three times in total, but stops immediately upon getting an even number. The diagram shows the outcomes and the corresponding prize amounts.
From each drop-down menu, select the option that creates the most accurate statement based on the information provided.
The probability that the player wins $150 is and the probability that the player wins at least $50 is .
For a player to win exactly $150, they must roll an even number on the first roll. The probability of this is \(\frac{1}{2}\).
To win at least $50, the player must avoid losing entirely. The only way to lose is by rolling an odd number on all three attempts, which has a probability of \((\frac{1}{2})^3 = \frac{1}{8}\). Therefore, the probability of winning at least $50 is \(1 - \frac{1}{8} = \frac{7}{8}\).
Correct answer:
Dropdown 1: "1/2"
Dropdown 2: "7/8"
Bunuel
A player rolls a fair six-sided die with faces numbered 1 through 6. If the result is even, the player wins a prize and does not roll again. If the result is odd, the player rolls again. The player may roll the die up to three times in total, but stops immediately upon getting an even number. The diagram shows the outcomes and the corresponding prize amounts.
From each drop-down menu, select the option that creates the most accurate statement based on the information provided.
The probability that the player wins $150 is and the probability that the player wins at least $50 is .
For a player to win exactly $150, they must roll an even number on the first roll. The probability of this is \(\frac{1}{2}\).
To win at least $50, the player must avoid losing entirely. The only way to lose is by rolling an odd number on all three attempts, which has a probability of \((\frac{1}{2})^3 = \frac{1}{8}\). Therefore, the probability of winning at least $50 is \(1 - \frac{1}{8} = \frac{7}{8}\).
Correct answer:
Dropdown 1: "1/2"
Dropdown 2: "7/8"
General Discussion
24 Oct 2024, 07:20
Kudos
Bookmarks
1. Probability to win $150
= 1/2
Since probability to get 150 in the first attempt is 1/2 [The question stem mentions that he is allowed to throw the dice three times or until he wins, whichever is earliest]
2. Probability to win at least $50 is
= 1/2 + (1/2 * 1/2) + (1/2 * 1/2 * 1/2)
= 1/2 + 1/4 + 1/8
= 7/8
He can win any of the three scenarios!
= 1/2
Since probability to get 150 in the first attempt is 1/2 [The question stem mentions that he is allowed to throw the dice three times or until he wins, whichever is earliest]
2. Probability to win at least $50 is
= 1/2 + (1/2 * 1/2) + (1/2 * 1/2 * 1/2)
= 1/2 + 1/4 + 1/8
= 7/8
He can win any of the three scenarios!
