Urn 1 contains 6 green and 5 yellow marbles Urn 2 contains 7

Urn 1 contains 6 green and 5 yellow marbles. Urn 2 contains 7 green and 9 yellow marbles. An experiment consists of choosing one of two urns at random then drawing a marble from the chosen urn. Urn 1 is more likely to be chosen than Urn 2 with probability 0.54.

What is the probability that a green marble was chosen, if it is known that Urn 2 was chosen?
a. 0.2354
b. 0.7778
c. 0.6512
d. 0.4375
e. 0.3845

A binomial experiment consists of four independent trials. The probability of success in each trial is ⁴³?₁₀₀ . Find the probabilities of obtaining exactly 0 successes, 1 success, 2 successes, 3 successes, and 4 successes, respectively, in this experiment.

a) [0, 0.1056, 0.3185, 0.3604, 0.1813]
b) [0.1056, 0.0074, 0.1802, 0.1813, 0.0342]
c) [0.1056, 0.0796, 0.4205, 0.0453, 0.0342]
d) [0.1056, 0.0796, 0.3604, 0.4216, 0.0342]
e) [0.1056, 0.3185, 0.3604, 0.1813, 0.0342]

Solution

1)

Urn 1 contains 6 green and 5 yellow marbles.

Urn 2 contains 7 green and 9 yellow marbles.

P(Urn 1) = 0.54

P(urn 2) = 0.46

probability that a green marble was chosen, if it is known that Urn 2 was chosen

P = Probaility green marble was chose / Urn 2 chosen

P(Green/Urn 2 ) = 7/16 = 0.4375

Option D

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2)

A binomial experiment consists of four independent trials.

P(S) = 0.43

P(F) = 0.57

a) 0 successes

P1 = 4c0*(0.57)^4 = 0.1055

b) 1 successes

P2 = 4c1*(0.43)*(0.57)^3 = 0.3185

c) 2 successes

P3 = 4c2*(0.43)^2*(0.57)^2 = 0.3604

d) 3 successes
P4 = 4c3*(0.43)^3*(0.57)^1= 0.1812

e) 4 successes
P5 = 4c4*(0.43)^4*(0.57)^0= 0.0341

Option (e) is correct

Hope this will help you !!!!!