Question

A Fair coin is tossed and a marble is chosen from the bag whose contents are shown below. what is the probability of the coin Landing heads up and a Blue Marble being randomly selected?
Blue: 3 Green: 4 Red: 3

A) 3/20
B) 1/6
C) 4/5
D) 5/6​

Answer 1

Answer: Option A)

$P=\frac{3}{20}$

Step-by-step explanation:

We know that when you toss the coin, the probability of getting heads up is

$p_m = 0.5$

In the bag there are 10 marbles

Blue: 3 Green: 4 Red: 3

So the probability of getting a blue marble is:

$P_b = \frac{3}{10}$

Note that the events are independent.

Finally, the probability of obtaining a head up and a blue marble will be:

$P = p_m * p_b$     → Because the events are independent:

$P = 0.5*\frac{3}{10}$

$P=\frac{3}{20}$

Answer 2

Answer:

P(Coin landing heads and blue marble is randomly selected) = 3/20.

Step-by-step explanation:

In this question, two events simultaneously take place. First, all the probabilities have to be identified. It is mentioned that the coin is a fair coin, therefore the probabilities of all the outcomes associated with the coin tossing will be equal. Therefore:

P(Coin landing heads) = 1/2.

P(Coin landing tails) = 1/2.

There are a total of 3 blue + 4 green + 3 red = 10 marbles in the bag. Therefore:

P(Selected marble is blue) = 3/10.

P(Selected marble is green) = 4/10.

P(Selected marble is red) = 3/10.

Assuming that both the events are independent, the probabilities of both the events can safely be multiplied. Therefore:

P(Coin landing heads and blue marble is randomly selected) = 1/2 * 3*10 = 3/20.

Therefore, the answer is 3/20!!!