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A bag contains 8 red and 5 white balls. Three balls are drawn at random. Find the probability that:i All the three balls are white.ii All the three balls are red.iii One ball is red and two balls are white.
Solution
Total number of balls = 8 + 5 = 13
Total number of events for drawing 3 balls = 13C3
(i) Total number of events for getting white balls = 5C3
P(all 3 balls white) = favourable outcomestotal outcomes=5C313C3
=5!3!×2!×3!×10!13!
⇒P(all 3 balls white)=5143
(ii) Favourable number of events for getting red balls = 8C3
P(all 3 balls red) = favourable outcomestotal outcomes=8C313C3
=8!3!×5!×3!×10!13!
⇒P(all 3 balls red)=28143
(iii) Favourable number of events for getting 1 red ball = 8C1
Favourable number of events for getting 2 white balls = 5C2
P(1 red and 2 white balls) = favourable outcomestotal outcomes=8C1×5C213C3
=8!1!×7!×5!2!×3!×3!×10!13!
⇒P(all 3 balls red)=40143
Mathematics
RD Sharma XI (2019)
All
2
Gate Exam Numerical Ability Probability
- (8c2+6c2)/14c2=43/91
- 10 years agoHelpfull: Yes(17) No(0)
- probability of drawing two balls of white colour is (8/14)*(7/13) = 56/182
probability of drawing two balls of red colour is (6/14)*(5/13) = 30/182
The probability of drawing two balls of same colour is 56/182 + 30/182 = 86/182 = 43/91
- 10 years agoHelpfull: Yes(16) No(5)
- ans=43/91
E(n)=8c2+6c2
s(n)=14c2
P(n)=E(n)/S(n)=43/91
- 10 years agoHelpfull: Yes(5) No(0)
- (8C2+6C2)/14C2=43/91
- 9 years agoHelpfull: Yes(0) No(0)
- c. 28/91+15/91
- 7 years agoHelpfull: Yes(0) No(0)
- 43/91, Its 8c2+6c2/14c2
- 6 years agoHelpfull: Yes(0) No(0)
- n(S)={14C2}= 91
n(E)={8C2+6C2}= 43
P(E)= 43/91
- 4 years agoHelpfull: Yes(0) No(0)