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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 .
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 .
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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 \(1 - \frac{1}{8} = \frac{7}{8}\).
I did not quite understand the solution. i am not able to understand 2nd part. please explain.
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You win at least $50 if you get even on any roll. You only lose if you roll odd all three times, which is (1/2)^3 = 1/8. So the chance of winning at least $50 is 1 - 1/8 = 7/8.