Question

Adam and Martha each performed a simulation. Adam's simulation He tossed a fair coin 4 times. Each time the coin landed with tails face up. Martha's simulation She randomly chose 1 block from each of two bags containing blocks. Each block she chose was labeled with the letter R. The probability for both Adam's and Martha's simulation were the same. Which two bags could be the ones Martha used in her simulation?

Answer 1

Answer:

Option D is correct.

B₁ = 4 and B₂ = 4

Step-by-step explanation:

The options of this question are attached.

Adam's simulation:

He tossed a fair coin 4 times and each time the coin landed with tails face up.

The probability that the coin lands with tail face up is

P(tails) = 1/2

Since there are two possible outcomes, heads or tails, so the chances of each outcome are 1/2.

It is given that Adam tossed the coin 4 times and each time got tails so the overall probability is

P(4 tails) = P(tails)*P(tails)*P(tails)*P(tails)

P(4 tails) = (1/2)*(1/2)*(1/2)*(1/2)

P(4 tails) = 1/16

Martha's simulation:

She randomly chose 1 block from each of two bags containing blocks. Each block she chose was labeled with the letter R.

So the probability is

P(2 R) = P(1 R)*P(1 R)

P(2 R) = (1/B₁)*(1/B₂)

Where B₁ and B₂ is the total number of blocks in each bag.

We are given that the probability for both Adam's and Martha's simulation were the same so,

1/16 = (1/B₁)*(1/B₂)

Refer to the attached options, only option D satisfies the above equation.

B₁ = 4 and B₂ = 4

1/16 = (1/4)*(1/4)

1/16 = 1/16   (satisfied)

None of the other option satisfies the above equation, Therefore, option D is correct.