Question

A bag of coins contains $10 in coins. 5 loonies ($1 coin), 10 quarters ($0.25), 20 dimes ($0.10) and the rest in nickels ($0.05). One coin is pulled at random from the bag, what is the probability it is a $1 coin?One coin is pulled at random, what is the probability it is a $0.10 or $0.05 coin?2 coins are pulled from the bag one at a time, but replaced in between the draw. What is the probability that both coins are dimes ($0.10)?2 coins are pulled from the bag one at a time, but the coin is not replaced. What is the probability that the first coin is a dollar ($1) and the second is a quarter ($0.25)?The probability of drawing 2 coins that have a value of exactly $1.25 is 0.05 or 100/1980. Why is this probability different then the probability calculated in d?

Answer 1

Given:

There is a bag of coins containing $10 in coins.

The types are,

5 loonies ($1 coin),

10 quarters ($0.25),

20 dimes ($0.10) and

The rest are in nickels ($0.05).

Let the number of coins in nickel is x.

So that,

`\begin{gathered} 5(1)+10(0.25)+20(0.10)+x(0.05)=10 \\ 5+2.5+2+0.05x=10 \\ 0.05x=0.5 \\ x=(50)/(5) \\ x=10 \end{gathered}`

Therefore, the number of nickel coins is 10.

The total number of coins is 45.

(i) To find the probability that a chosen coin is $1 coin:

So, the answer is

`\begin{gathered} P(A)=(5)/(45) \\ =(1)/(9) \end{gathered}`

(ii) To find the probability that a chosen coin is $0.10 or $0.05 coin:

So, the answer is,

`\begin{gathered} P(B)=(20)/(45)+(10)/(45) \\ =(30)/(45) \\ =(2)/(3) \end{gathered}`

(iii) To find the probability that both the two chosen coin dimes coin( with replacement):

So, the answer is,

`\begin{gathered} P(C)=(20)/(45)*(20)/(45) \\ =(16)/(81) \end{gathered}`

(iv) To find the probability that the first coin is a dollar ($1) and the second is a quarter ($0.25): (without replacement)

So, the answer is,

`\begin{gathered} P(D)=(5)/(45)*(10)/(44) \\ =(5)/(198) \end{gathered}`

(v) To find the probability of drawing 2 coins that have a value of exactly $1.25:

So, the value is,