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What is the probability that you will first cut an ace

a) on the 5th cut ?
b) in fewer than 4 cuts?
c) what is the expected waiting time before you cut an ace ?

User Jerclarke
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1 Answer

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Final answer:

The probability of cutting an ace on the 5th cut is calculated by multiplying the probability of not getting an ace in the first four cuts with the probability of getting an ace on the fifth cut. The probability of cutting an ace in fewer than 4 cuts is found by subtracting the probability of not getting an ace in the first three cuts from 1. The expected waiting time before cutting an ace is calculated using the geometric distribution, giving an average of 13 cuts.

Step-by-step explanation:

The student has asked about the probability of cutting an ace from a deck of cards at various instances and the expected waiting time before cutting an ace. The subject is Mathematics at High School level. Here's how to calculate each part of the question:

a) Probability of first cutting an ace on the 5th cut

There is a probability of 48/52 for not getting an ace each time we make a cut for the first four cuts (assuming the deck is shuffled after each cut). Then on the fifth cut, there is a 4/52 chance of getting an ace. So, the probability of first cutting an ace on the 5th cut is:

(48/52) * (48/52) * (48/52) * (48/52) * (4/52)

b) Probability of cutting an ace in fewer than 4 cuts

This is the complement of not getting an ace in the first three cuts. So, calculate the probability of not getting an ace in any of the first three cuts (48/52) for each cut and subtract this from 1:

1 - ( (48/52) * (48/52) * (48/52) )

c) Expected waiting time before you cut an ace

The expected waiting time can be calculated using the geometric distribution. For a deck of cards, the probability of success (cutting an ace) on a given cut is 4/52. The expected number of trials (cuts) before the first success is 1/(success probability), which is:

1/(4/52) = 52/4 = 13 cuts

Therefore, on average, you would expect to cut an ace every 13 cuts.

User Qxotk
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