Question

please help..... would mark u brainliest

Answer 1

It is given that "the probability that the face value is odd is 90% of the probability that the face value is even".
P(1) + P(3) + P(5) = 0.9 * (P(2) + P(4) + P(6)) → equation (1)

Also, "The probability of getting any even numbered face is the same."
So, P(2) = P(4) = P(6) = x → equation (2)

from equations (1) & (2),
P(1) + P(3) + P(5) = 0.9 * (3x) = 2.7x → equation (3)

We know that the total probability of sample space = 1
i.e., P(1) + P(2) + P(3) + P(4) + P(5) + P(6) = 1
(P(1) + P(3) + P(5)) + (P(2) + P(4) + P(6)) = 1
2.7x + 3x = 1 ⇒ x = 0.17544 = P(2) = P(4) = P(6)

Probability that the face is even given that it is greater than 3 is 0.75.

P(4)+P(6)P(4)+P(5)+P(6)=0.75

From above equation, The probability that the face value exceeds 3 = P(4)+P(5)+P(6)=P(4)+P(6)0.75=2∗0.175440.75= 0.46784