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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