Question

Urn A has 12 white and 8 red balls. Urn B has 2 white and 17 red balls. We flip a fair coin. If the oucome is heads, then a ball from urn A is selected, whereas if the oucome is tails, then a ball from urn B is selected. Suppose that a red ball is selected. What is the probability that the coin landed heads?

Answer 1

\[P(A|B)=\frac{P(B|A)P(A)}{P(B|A)P(A)+P(B|A^c)P(A^c)}\] put A = coin is heads, B = red ball is selected and compute

Answer 2

the numerator will be
\[\frac{17}{19}\times \frac{1}{2}\]

Answer 3

the denominator will be
\[\frac{17}{19}\times \frac{1}{2}+\frac{8}{20}\times \frac{1}{2}\]

Answer 4

Should that be the final answer? The answer of the numerator over the answer of the denominator?

Answer 5

@robtobey do you think you know how to solve this?

Answer 6

Probability of red ball given heads:\[\frac{1}{2}\frac{8}{20}=\frac{1}{5} \]Probability of red ball given tails:\[\frac{1}{2}\frac{17}{19}=\frac{17}{38} \]Probability of heads given a red ball was drawn:\[\frac{\frac{1}{5}}{\frac{1}{5}+\frac{17}{38}}=\frac{38}{123} \]

Answer 7

@satellite73 may have the wrong numerator.

Answer 8

Thank you!!

Answer 9

Do you think you could help me solve this one as well? It seems like an easy method but I can't seem to get it right.


Consider two people being randomly selected. (For simplicity, ignore leap years.)
(a) What is the probability that two people were born on Monday?

Answer 10

\[\frac{1}{7}\frac{1}{7}=\frac{1}{49} \]Let A be the first person and B be the second person. The following is the probability of A being Monday born and B not, A not and B Monday born, A and B not.\[\left(\frac{1}{7}\frac{6}{7}+\frac{6}{7}\frac{1}{7}+\frac{6}{7}\frac{6}{7}\right)=\frac{48}{49} \]1- the above is the probabiliy that both were born on Monday.\[1-\frac{48}{49}\text{=}\frac{1}{49}\]

Answer 11

you are right. red ball given heads is \(\frac{8}{20}=\frac{2}{5}\)

Answer 12

THANK YOU :)

Answer 13

sorry i messed up
robtoby answer of \(\frac{38}{123}\) is hte right one