A bag of marbles contains 5 red marbles, 5 green marbles, 5 yellow marbles and 5 blue marbles. If Mya takes one marble at a time at random, what is the probability that she takes out a red marble the first time and a green marble the second?

The probability that she takes out a red marble the first time and a green marble the second

P(E₁ ∩ E₂ ) = P(E₁) P(E₂)

Step-by-step explanation:

Step(i):-

Given a bag of marbles contains 5 red marbles, 5 green marbles, 5 yellow marbles and 5 blue marbles

Total balls n(S) = 5red +5green +5yellow + 5blue  =  20

Total numbers cases

                         n(S) = 20C₁ = 20 ways

Step(ii):-

Let E₁ be the event of drawing one red marble from 5 red balls

                       n(E₁) = 5C₁

Let E₂ be the event of drawing one green marble from 5 green balls

                       n(E₂) = 5C₁

The two events are independent events

P(E₁ ∩ E₂ ) = P(E₁) P(E₂)

                =

Final -

 The probability that she takes out a red marble the first time and a green marble the second

P(E₁ ∩ E₂ ) = P(E₁) P(E₂)

                =

the answer is simply just 6n add the two together as they have an n