Question

A bag contains 2 blue marbles, 3 red marbles, and 5 yellow marbles. What are the odds of randomly pulling a marble, then pulling out another without replacing the first, and having both be red marbles

Answer 1

`Pr(picking\text{ both reds) = }(1)/(15)\text{ (option A)}`

Step-by-step explanation:

number of blue marbles = 2

number of red marbles = 3

number of yellow marbles = 5

Total marbles = 2 + 3 + 5 = 10

Probability of picking both reds = probability of picking red as 1st and red as second

Probability of picking a red marble = number of red marbles/ total marbles

`Pr(\text{red marbles) = }(3)/(10)`

Probability of picking red as 1st = 3/10

For the 2nd picking:

Since it is a probability without replacement (after picking the 1st one, it was not replace), the number of red marbles will reduce by 1

number of red amrbles = 3 - 1 = 2

The total number of marbles will also decrease by 1

Total marbles = 10 - 1 = 9

Probability of picking red as 2nd = 2/9

`\begin{gathered} Pr(picking\text{ both reds) = }(3)/(10)\text{ }*\text{ }(2)/(9) \\ Pr(picking\text{ both reds) = }(1)/(15)\text{ (option A)} \end{gathered}`