Q4. - Algebra 1

Question

One coin is randomly selected from a jar containing 70 nickels, 100 dimes, 80 quarters, and 50 one-dollar coins. Find each probability.

P (value greater than $0.10)

Step-by-Step Solution

Answer

The probability of choosing a value greater than $0.10 is $\frac{13}{30}$.

Step 1. State the formula.

The formula for the probability of an event is $P\left(E\right)=\frac{n}{T}$, where E denotes the event, P(E) denotes the probability of the event E, n denotes the number of favorable outcomes and T denotes the total number of outcomes.

Step 2. List the given data.

The jar contains 70 nickels, 100 dimes, 80 quarters, and 50 one-dollar coins.

Step 3. Explanation.

The event is “choosing a value greater than $0.10.”

 

The jar contains 70 nickels, 100 dimes, 80 quarters, and 50 one-dollar coins.

 

The value of a nickel is $0.05.

 

The value of a dime is $0.10.

 

The value of a quarter is $0.25.

 

Finally, the value of a one-dollar coin is obviously $1.

 

So, the event “choosing a value greater than $0.10” is equivalent to the event “choosing a quarter or one-dollar coin”.

 

Since the jar contains 80 quarters and 50 one-dollar coins,

$$\begin{gathered} n=80+50 \\ =130 \end{gathered}$$

So, $n=130$.

 

The total number of coins in the jar is $70+100+80+50=300$.

 

Then, $T=300$.

Step 4. Calculate the probability.

Put E : value greater than $0.10, $n=130$ and $T=300$ in $P\left(E\right)=\frac{n}{T}$ to get,

$$\begin{gathered} P\left(\mathrm{value} \mathrm{greater} \mathrm{than} \$0.10\right)=\frac{130}{300} \\ =\frac{13}{30} \end{gathered}$$
 

So,  $P\left(\mathrm{value} \mathrm{greater} \mathrm{than} \$0.10\right)=\frac{13}{30}$

 

The probability of choosing a value greater than $0.10 is $\frac{13}{30}$.