CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

(i) An urn contains 3 white, 4 red and 5 black balls. Two balls are drawn one by one without replacement. What is the probability that at least one ball is black?
(ii) A bag contains 4 white, 7 black and 5 red balls. Three balls are drawn one after the other without replacement. Find the probability that the balls drawn are white, black and red respectively.

Open in App
Solution

(i) Consider the given events.
A = A white or red ball in the first draw
B = A white or red ball in the second draw

Now, PA=712PB/A=611 PAB=PA×PB/A =712×611 =722 Required probability = 1-PAB =1-722 =1522

(ii) Consider the given events.
A = A white ball in the first draw
B = A black ball in the second draw
C = A red ball in the third draw

Now, PA=416=14PB/A=715PC/AB=514 Required probability = PABC=PA×PB/A×PC/AB =14×715×514 =124

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