suppose we have two hats one has 4 red balls and 6 green balls, the other has 6 red and 4 green. we toss a fair coin, if heads, pick a random ball from the first hat, if tails from the second. What is the probability of getting a red ball.
Solution:
P(Red)=P(Red∣H).P(H)+P(Red∣T).P(T)=410.12+610.12=12P(Red)=P(Red|H).P(H)+P(Red|T).P(T) \\=\dfrac{4}{10}.\dfrac{1}{2}+\dfrac{6}{10}.\dfrac{1}{2} \\=\dfrac{1}{2}P(Red)=P(Red∣H).P(H)+P(Red∣T).P(T)=104.21+106.21=21
Need a fast expert's response?
and get a quick answer at the best price
for any assignment or question with DETAILED EXPLANATIONS!
Comments
Leave a comment