Afair 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

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!!!

Option A)

Step-by-step explanation:

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

In the bag there are 10 marbles

Blue: 3 Green: 4 Red: 3

So the probability of getting a blue marble is:

Note that the events are independent.

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

    → Because the events are independent:

area = length x width

length = 2 x width

32 = 2 x width x width

32 = 2 x width^2

36 = width ^2  

width = 6

length = 12

area is 32

perimeter = 2 x 12 + 2 x 6 = 24 + 12 = 36

step-by-step explanation:

its a straight line parallel to the time axis