Question

Laura has a bag with four red marbles and six blue marbles. She randomly pulls a red marble out, replaces it, and randomly pulls another red marble out. She believes that she can calculate the probability of this happening by doubling the probability of pulling a red marble out on the first try. Explain why this incorrect.

Answer 1

Write out the formula of Probability

`\text{Probability}=\frac{Number\text{ of favourable outcomes}}{Total\text{ number of outcomes}}`

Total number of marbles= total number of red marbles plus the total number of blue marbles.

Total number of marbles= 4+6= 10 marbles.

We were the two different marbles were picked and replaced randomly. So let us get the probability of picking a red marble.

Doubling the probability of pulling a red marble out on the first try is not correct, it will be shown why that is not correct shortly below.

`\begin{gathered} \text{Number of red marbles=4} \\ \text{Total marble= 10} \\ \text{Probability of red marble=}(4)/(10) \\ =\frac{\text{ 2}}{5} \end{gathered}`
`\begin{gathered} \text{Probability of picking another red will also be}(2)/(5)\text{ because the red marble} \\ is\text{ b}een\text{ repolaced.} \end{gathered}`

Therefore, the probability of pulling a red marble and also another red marble will result to:

`(2)/(5)*(2)/(5)=(4)/(25)`

But if you just double the red marble on your first try, you will have:

`\begin{gathered} 2*(2)/(5)=(4)/(5) \\ \text{which is not correct} \end{gathered}`

From the calculations done above you can see doubling it is not correct.