What is the Set and Probability of Drawing Coins with Replacement?

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In summary, the set S for the random experiment of drawing two coins with replacement from a coin purse containing 9 nickels and 1 quarter is {two nickels, two quarters, one nickel and one quarter}. The probabilities of the elementary events are P[two nickels] = 45/55, P[two quarters] = 3/55, P[one nickel and one quarter] = 7/55. The probability of drawing a quarter for the first coin is 1/10, and with replacement, the probability for the second coin would also be 1/10. The probability of drawing two quarters would be 1/100. The probabilities for drawing two nickels and one nickel and one quarter are
  • #1
magnifik
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A coin purse contains 9 nickels and 1 quarter. Two coins are drawn WITH REPLACEMENT. What is the set S of this random experiment? What are the probabilities of the elementary events?

i'm a little confused by the "with replacement" concept. I'm just wondering if I'm doing it correctly. i have the equation n-1+k choose n-1 so here's my work:

set S = {two nickels, two quarters, one nickel and one quarter}
total possible ways of selection: (10-1+2) choose (10-1) = 55
P[two nickels] = (9-1+2) choose (9-1) / 55 = 45/55
P[two quarters] = (2-1+2) choose (2-1) / 55 = 3/55
P[one nickel and one quarter] = 1 - Pall other events = 1 - (45/55 + 3/55) = 7/55

have i done the problem correctly?
 
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  • #2
What is the probability of drawing a quarter for the first coin? With replacement, would the probability of drawing a quarter the second time be any different?

The results of the coin drawings would be independent events, so what would the probability of drawing two quarters be, in terms of the probabilities of drawing a quarter for the first and second drawings?
 
  • #3
MisterX said:
What is the probability of drawing a quarter for the first coin? With replacement, would the probability of drawing a quarter the second time be any different?

The results of the coin drawings would be independent events, so what would the probability of drawing two quarters be, in terms of the probabilities of drawing a quarter for the first and second drawings?

probability for drawing a quarter for the first coin = 1/10
probability of drawing a quarter for the second coin = 1/10
so P[drawing 2 quarters] = 1/100 ?

do i follow the same approach for the probability of each of the elementary events? like for 1 nickels..
probability of drawing 1 nickel first = 9/10
probability of drawing 1 nickel second = 9/10
so P[drawing 2 nickels] = 81/100

and then for 1 nickel and 1 quarter
probability of drawing 1 nickel = 9/10
probability of drawing 1 quarter = 1/10
P[1 nickel, 1 quarter] = 9/100

but the probabilities do not add up to 1 so i am still confused :\
 
  • #4
Mod note: Moved from Engineering & CS section.
 
  • #5
magnifik said:
but the probabilities do not add up to 1 so i am still confused :\
There are two ways the two coins drawn would be a nickel and a quarter.
 
Last edited:
  • #6
MisterX said:
There are two ways the two coins drawn would be a nickel and a quarter.

Ahh, I see. This is because order doesn't matter, correct? so P[1 nickel, 1 quarter] = 9/100 + 9/100 = 18/100 = 9/50
 

FAQ: What is the Set and Probability of Drawing Coins with Replacement?

What is "probability with replacement"?

"Probability with replacement" refers to a type of probability calculation where an event or outcome can occur multiple times in a series of trials, and each time it occurs, it is considered independent from the previous occurrences.

How is "probability with replacement" different from "probability without replacement"?

In "probability without replacement", an event or outcome can only occur once in a series of trials, and each occurrence affects the probability of future occurrences. In "probability with replacement", the likelihood of an event or outcome remains the same regardless of previous occurrences.

Can "probability with replacement" be applied to real-life situations?

Yes, "probability with replacement" can be applied to real-life situations such as rolling a die, drawing cards from a deck, or flipping a coin multiple times. It is important to note that in real-life situations, the probability may not be exactly the same due to factors such as wear and tear or human error.

How do you calculate the probability with replacement?

The probability with replacement can be calculated by dividing the number of times the event or outcome can occur by the total number of possible outcomes. For example, if you roll a six-sided die and want to calculate the probability of rolling a 3, you would divide 1 (since there is only one 3 on the die) by 6 (the total number of possible outcomes).

Are there any limitations to using "probability with replacement"?

One limitation of using "probability with replacement" is that it assumes each trial is independent from the previous ones, which may not always be the case in real-life situations. Additionally, the accuracy of the probability calculation may be affected by factors such as sample size and the randomness of the events or outcomes.

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