Question

6. Carissa has a large bag of coins. After counting the coins, she recorded the counts in the table below. She then decided to draw some coins at random, replacing each coin before the next draw.QuartersNickelsDimesPennies25154018(a) What is the probability that Carissa obtains a quarter on the first draw? (b) What is the probability that Carissa obtains a penny or a dime on the second draw? (c) What is the probability that Carissa obtains at most 10 cents worth of money on the third draw? (d) What is the probability that Carissa does not get a nickel on the fourth draw? (e) What is the probability that Carissa obtains at least 10 cents worth of money on the fifth draw?

Answer 1

Given

The total is 98

Part A

`\text{Probability of OBTAINING a quarters= }(25)/(98)`

Part B

`\text{Probability of obtaining dimes or pennies =}(18)/(98)+(40)/(98)=(58)/(98)=(29)/(49)`

Part C

Recall

`\begin{gathered} 25\text{cent = 1 Quarter} \\ 5\text{cent =Nickel} \\ 10\text{cent =dime} \\ 1\text{ cent = 1penny} \end{gathered}`

The probability of at most 10 cents

`\therefore\text{ The probability of others =}(15+40+18)/(98)=(73)/(98)`

Part D

`\begin{gathered} Probability\text{ of not obtaining Nicke i.}e\text{ substract nickel from total} \\ Probability\text{ of not obtaining Nicke=}(98-15)/(98)=(83)/(98) \end{gathered}`

Part E

We can obtain 10cents five times from the given Quarters and Dimes

since 25quarter =625

`\begin{gathered} \text{Probability of getting at least 10 cents on the fifth } \\ i\mathrm{}e\text{ it is only possible on Quarter and Dimes} \\ \\ \text{Probability of getting at least 10 cents on the fifth }=(40+25)/(98)=(65)/(98) \end{gathered}`