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!!!
Answer: Option A)
[tex]P=\frac{3}{20}[/tex]
Step-by-step explanation:
We know that when you toss the coin, the probability of getting heads up is
[tex]p_m = 0.5[/tex]
In the bag there are 10 marbles
Blue: 3 Green: 4 Red: 3
So the probability of getting a blue marble is:
[tex]P_b = \frac{3}{10}[/tex]
Note that the events are independent.
Finally, the probability of obtaining a head up and a blue marble will be:
[tex]P = p_m * p_b[/tex] → Because the events are independent:
[tex]P = 0.5*\frac{3}{10}[/tex]
[tex]P=\frac{3}{20}[/tex]
