Probability of Prevesh Paying Three Days in a Row: Coin Toss Lunch Break

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The discussion centers on calculating the probability of Pravesh paying for lunch three consecutive days based on a coin toss. The probability of Pravesh paying on any given day is 1/2, leading to a cumulative probability of 1/8 for him paying three days in a row. Participants debated the correct answer, with one suggesting that the probability is 0.125, derived from multiplying 1/2 three times. There is also a reminder against doing homework for others, emphasizing the importance of independent problem-solving. The consensus indicates that the probability of Pravesh paying three days in a row is indeed 0.125.
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Pravesh and Solly usually go to a coffee shop during their lunch break and toss a coin to see who will pay. The probability that Prevesh will pay three days in a row is

(1) 0.025
(2) 0.500
(3) 0.125
(4) 0.015

My attempt (I am not sure if this is right)

X= number of days that Prevesh will pay

x- P(x)
0 - 0.250
1 - 0.250
2 - 0.250
3 - 0.250

So I assume that the correct answer is (1)?
 
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i think since pravesh would have 1/2 chance of paying for a lunch on one day, you would have to multiply that by 3 and the odds become less for him to pay 3 times in a row therefore you would get 1/8 from 1/2 * 1/2 *1/2. Getting 0.125
 
Thank you
 
Master7731 said:
i think since pravesh would have 1/2 chance of paying for a lunch on one day, you would have to multiply that by 3 and the odds become less for him to pay 3 times in a row therefore you would get 1/8 from 1/2 * 1/2 *1/2. Getting 0.125

Please do not do the OP's homework for him. It is cheating, and not allowed here at the PF.
 
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