Suppose a game show called Bayesian Boxes has the contestant sample one marble from a box and guess what type of box it is. When the contestant picks a box, the chance of each box type for this round is as follows: Type A: 75%, Type B: 20%, and Type C: 5% The chance of drawing a red marble from Type A is 2%, from Type B is 25%, and from Type C is 98%. The contestant selects a box and draws a marble. What is the chance of drawing a red m

Question

contestada

Respuesta :

Tundexi Tundexi · Answer 1 · 2020-09-13T03:15:59+02:00

Answer:

P(C/R) = 0.4298

Step-by-step explanation:

Missing word "arble, what is the chance it came from box Type C? (Give the decimal answer rounded to at least four places )"

Solution:

Given that P(A) = 75% = 0.75

P(B) = 20% = 0.20

P(C) = 0.05

The chance of drawing a Red marble fro,

P(R/A) = 2% = 0.02

P(R/B) = 25% = 0.25

P(R/C) = 98% = 0.98

If he select Box A than the probability of Red marble drawing

= P(A) * P(R/A)

= 0.75 * 0.02

= 0.015

If he select Box B than the probability of Red marble drawing

= P(B) * P(R/B)

= 0.20 * 0.25

= 0.5

If he select Box C than the probability of Red marble drawing

= P(C) * P(R/C)

= 0.98 * 0.05

= 0.049

So total probability of drawing Red marble is

P(R) = 0.015 + 0.050 + 0.049

P(R) = 0.114

If the Contestant draw a Red marble. Probability that it comes from box type (C) using Boyes theorem is

P(C/R) = {P(C) * P(R/C)} / P(R)

P(C/R) = 0.049 / 0.114

P(C/R) = 0.429824

P(C/R) = 0.4298