If I chose a number uniformly from the integers from 1 to 25, calculate the conditional probability that the number is a multiple of 6 (including 6) given that it is larger than 18.

Step-by-step explanation:

If I choose a number from the integers 1 to 25, the total number of integers I can pick is the total outcome which is 25. n(U) = 25

Let the probability that the number chosen at random is a multiple of  6 be P(A) and the probability that the number chosen at random is is larger than 18 be P(B)

P(A) = P(multiple of 6)

P(B) = P(number larger than 18)

A = {6, 12, 18, 24}

B = {19, 20, 21, 22, 23, 24, 25}

The conditional probability that the number is a multiple of 6 (including 6) given that it is larger than 18 is expressed as P(A|B).

P(A|B) = P(A∩B)/P(B)

Since probability = expected outcome/total outcome

A∩B = {24}

n(A∩B) = 1

P(A∩B) = n(A∩B)/n(U)

P(A∩B) = 1/25

Given B = {19, 20, 21, 22, 23, 24, 25}.

n(B) = 7

p(B) = n(B)/n(U)

p(B) = 7/25

Since P(A|B) = P(A∩B)/P(B)

P(A|B) = (1/25)/(7/24)

P(A|B) = 1/25*25/7

P(A|B) = 1/7

Hence the conditional probability that the number is a multiple of 6 (including 6) given that it is larger than 18 is 1/7

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step-by-step explanation:

im eatin up the 44

1^3 + 2^3 + 3^3 +4^3 = (1+2+3+4)^2

step-by-step explanation:

i am only able to answer the first part. i am not sure about the second part. sorry