1) A bag contains 4 black, 5 blue, 6 green balls. 2 balls are drawn at random, what is the probability that it is black, blue ?
Solution:
Total number of balls=4+5+6=15
Let S be the sample space.
Then, n(S) = Total number of ways of drawing 2 balls out of 15
:.n(S) = {::}^15C_2=(15*1414*13!)/((2*1)*(13!))=(15*1414)/(2*1)=105
Let E= Event of drawing 2 black balls out of 4 or 2 blue balls out of 5.
n(E) = {::}^4 C_2 + {::}^5 C_2
=(4*33*2!)/((2*1)*(2!)) + (5*44*3!)/((2*1)*(3!))
=6 + 10
=16
:.P(E)=(n(E))/(n(S))=16/105
2) A bag contains 4 black, 5 blue, 6 green balls. 1 balls are drawn at random, what is the probability that it is not black, blue ?
Solution:
Total number of balls=4+5+6=15
Let S be the sample space.
Then, n(S) = Total number of ways of drawing 1 balls out of 15
:.n(S) = {::}^15C_1 [:' {::}^n C_1 = n]=15
Let E= Event of drawing 1 black balls out of 4 or 1 blue balls out of 5.
n(E) = {::}^4 C_1 + {::}^5 C_1
=4 + 5
=9
:.P(E)=(n(E))/(n(S))=9/15=3/5
:.P(E')=1-3/5=2/5
3) A bag contains 4 black, 5 blue, 6 green balls. 2 balls are drawn at random, what is the probability that it is same ?
Solution:
Total number of balls=4+5+6=15
Let S be the sample space.
Then, n(S) = Total number of ways of drawing 2 balls out of 15
:.n(S) = {::}^15C_2=(15*1414*13!)/((2*1)*(13!))=(15*1414)/(2*1)=105
Let E= Event of drawing same.
n(E) = {::}^4 C_2 + {::}^5 C_2 + {::}^6 C_2
=(4*33*2!)/((2*1)*(2!)) + (5*44*3!)/((2*1)*(3!)) + (6*55*4!)/((2*1)*(4!))
=6 + 10 + 15
=31
:.P(E)=(n(E))/(n(S))=31/105
This material is intended as a summary. Use your textbook for detail explanation.
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