A new surgical procedure is said to be successful 60% of the time. Suppose the operation is performed five times and the results
1 answer:
Answer:
0.07776, 0.3456,0.08704
Step-by-step explanation:
Given that a new surgical procedure is said to be successful 60% of the time
Let X be the no of successes
Then X has only two outcomes and each trial is independent of the other
Hence X is binomial with p = prob of success in a single trial = 0.60
n = Number of operations performed = 5
Prob that
a) all five operations are successful =
(b) exactly three are successful
=
(c) less than two are successful
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Answer:
The probability that the final settlement amount is between 1 and 3 given that the initial claim is 2 = (2/9) = 0.2222
Step-by-step explanation:
The complete question is presented in the attached image to this solution
The joint probability distribution is given as
f(x, y) = {2/[x²(x - 1)} × y^-[(2x-1)/(x-1)] for x>1 And y>1
Given that the initial claim estiamted by the comapny is 2, determine the probability that the final settlement amount is between 1 and 3.
That is, x = 2, and y ranges from 1 to 3
Inserting x = 2 into the expression, we obtain
f(y) = (1/2) × y⁻³ = (y⁻³/2)
The required probability would then be
P(1 < y ≤ 3) = ∫³₁ f(y) dy
= ∫³₁ (y⁻³/2) dy
= [y⁻²/-4]³₁
= [3⁻²/-4] - [1⁻²/-4]
= (-1/36) - (-1/4)
= (1/4) - (1/36)
= (8/36)
= (2/9) = 0.2222
Hope this Helps!!!
