Question

A jar has two nickels, three quarters, and a half-dollar coin in it. Three coins are randomly drawn.

1. (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.
2. (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.
3. (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.
4. (1) What is the probability that a selection of three coins includes a half-dollar coin?
5. (1) What is the probability that the selection of three coins does not include a half-dollar coin?
6. (2) What is the probability that the selection of three coins is worth less than $1?
7. (2) What is the probability that the selection of three coins is equal to $1?
8. (2) What is the probability that the selection of three coins is greater than $1?
9. (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 frequency	Cumulative P(X < x)
$0.35
$0.55
$0.60
$0.75
$1

Answer 1

All possible types of coins are: N, n, Q, q, p, H.

• Samples of three can be chosen in: { (N,n,Q), (N,n,q), (N,n,p), (N,n,H), (N,Q,q), (N,Q,p), (N,Q,H), (N,q,p), (N,q,H), (N,p,H), (n,Q,q), (n,Q,p), (n,Q,H), (n,q,p), (n,q,H), (n,p,H), (Q,q,p), (Q,q,H), (Q,p,H), (q,p,H) }.
  
• We consider each coins are distinct. There are 6 different kinds of coins. From the 6 coins, 3 coins can be chosen in
  6 3
  ways. Hence, combination is used to choose the coins. Permutation will not be used, as arrangements are not taken into consideration.
  
  
• From the 6 coins, Half a dollar coin can be selected in 1 ways.
  The remaining 2 coins from the 5 coins can be selected in $\binom{5}{2}$ ways.
  Hence, the required probability is $=\frac{\binom{5}{2}*1}{\binom{6}{3}}=0.5$
  
  
• Without the half a dollar coin, there are 5 coins.
  Thus, from the remaining 5 coins, three coins can be selected in $\binom{5}{3}$ ways.
  Hence, the required probability is
  من بن * 1 = 05 2.
  .
  
  
• For the sum of the values of the coin to be less than $1, the combinations of one half cent and two quarter cents should not be taken into account.
  The half dollar coin can be chosen in 1 way and the two quarter dollar coin can be chosen in $\binom{3}{2}$ ways.
  Hence, the required probability is $=\frac{\binom{6}{3}-\binom{3}{2}*1}{\binom{6}{3}}=0.85$ .
  
  
• For the sum of the values of the coin to be equal to $1, only the combinations of one half cent and two quarter cents should be taken into account.
  The half dollar coin can be chosen in 1 way and the two quarter dollar coin can be chosen in $\binom{3}{2}$ ways.
  Hence, the required probability is $=\frac{\binom{3}{2}*1}{\binom{6}{3}}=0.15$ .
  
  
• It is observed that, P(sum of amount is less than $1)+P(sum of amount is equal to $1) =1.
  Hence,
  P(sum of amount is greater than $1)
  = 1 - P(sum of amount is less than or equal to $1)
  = 1 - (P(sum of amount is less than $1)+P(sum of amount is equal to $1))
  = 1 - 1
  = 0.
  
  

As per HOMEWORKLIB POLICY, we are to answer the first four subparts. However, I have tried answering as many questions as I could in the given time limit.

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