Answer:
Option D is correct.
B₁ = 4 and B₂ = 4
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.