Question

10. A jar has two nickels, three quarters, and a half-dollar coin in it. Three coins are randomly drawn. a. (4) List the simple events in the sample space if order does not matter. You may use a code such as N= nickel 1, n= nickel 2, Q = quarter 1, q = quarter 2, p = quarter 3, and H = the half-dollar coin. b. (2) Show how to use a permutation or a combination formula to determine the number of options that could occur in the sample space if order does not matter. C. (2) Show how to use a permutation or a combination formula to determine the number of options that could occur in the sample space if order does matter. d. (1) What is the probability that a selection of three coins includes a half-dollar coin? e. (1) What is the probability that the selection of three coins does not include a half-dollar coin? f. (2) What is the probability that the selection of three coins is worth less than $1? g. (2) What is the probability that the selection of three coins is equal to $1? h. (2) What is the probability that the selection of three coins is greater than $1? i. (4) Draw or create a cumulative frequency table for the ways in which the sum of the coins can be shown. Hint: You want to show $0.35, $0.55, $0.60, $0.75, $1, etc. for the x column. You may add or delete rows to the table below. X frequency P(x) Cumulative P(X<x) Cumulative frequency $0.55 $0.60 $0.75 $1

Answer 1

a.If three coins are randomly drawn, then events in the sample spacei if order doesn't matter would be:

6C3= 20

NnQ,Nnq,Nnp,NnH,NQq,NQp,NQH,Nqp,NqH,NpH,nQq,nQp,nQH,nqp,nqH,npH,Qqp,QqH,QpH,qpH

b. I've already demonstrated but to be xplicit, if order of the coins picked doesn't matter, then it's a simple problem of choosing 3 coins from a total of 6 coins; for choosing when order is not important, we use combination, hence we use 6C3= 6!/(3!3!)=20

c. If order does matter, then whether a nickel is picked first or a quarter etc would matter. In these cases, where we have to both choose and the order is also important, we use permutation. So we use 6P3= 6!/3! = 120

Notice that the cases become 6 times the cases when ordering was not important, as each coin's place in the solution will give rise to 6 different cases for each of the 20 former cases. Hence 6*20=120

d. The probability that the selection of 3 coins includes a half dollar sign

So there will be two cases, when ordering is important and when it's not

d. i When ordering is not important

The total number of cases is 20

If we fix 1 position for half dollar coin, we are left with 5 coins out of which any 2 can be chosen,

hence possible cases = 5C2 = 10

So probability= 10/20=0.5=50%

d ii. When ordering is important, total nmber of cases are 120

If the first pick is a half dollar coin, then 5P2 cases arise i.e. 20 cases

if it's the second pick, against 20 cases, and if it's the third pick, again 20 ,

so total number of possible cases is 20*3=60

So the probability comes out to be 60/120=50%

So, order or no order, there is a 50% chnace that the 3 coins include a half coin.

e. probability for non selection of half coin is 1-prob that half coin is selected (0.5)= 0.5, 50% again