wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

The contents of three urns are as follows:
Urn 1 : 7 white, 3 black balls, Urn 2 : 4 white, 6 black balls, and Urn 3 : 2 white, 8 black balls. One of these urns is chosen at random with probabilities 0.20, 0.60 and 0.20 respectively. From the chosen urn two balls are drawn at random without replacement. If both these balls are white, what is the probability that these came from urn 3?

Open in App
Solution

Let E1, E2 and E3 denote the events of selecting Urn I, Urn II and Urn III, respectively.

Let A be the event that the two balls drawn are white.

PE1=20100 PE2=60100 PE3=20100Now,PA/E1=7C210C2=2145PA/E2=4C210C2=645PA/E3=2C210C2=145Using Bayes' theorem, we getRequired probability = PE3/A=PE3PA/E3PE1PA/E1+ PE2PA/E2++ PE3PA/E3 =20100×14520100×2145+60100×645+20100×145 =121+18+1=140


flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Conditional Probability
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon