4) A bag contains 4 black, 5 blue, 6 green balls. 3 balls are drawn at random, what is the probability that it is not 1 black, 2 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 `3` balls out of `15`
`:.n(S) = {::}^15C_3=(15*14*131314*1313*12!)/((3*2*1)*(12!))=(15*14*131314*1313)/(3*2*1)=455`
Let `E=` Event of drawing 1 black balls out of 4 and 2 blue balls out of 5.
`n(E) = {::}^4 C_1 xx {::}^5 C_2`
`=4 xx (5*44*3!)/((2*1)*(3!))`
`=4 xx 10`
`=40`
`:.P(E)=(n(E))/(n(S))=40/455=8/91`
`:.P(E')=1-8/91=83/91`
5) A bag contains 4 black, 5 blue, 6 green balls. 4 balls are drawn at random, what is the probability that it is 2 black, 2 other ?
Solution:
Total number of balls`=4+5+6=15`
Let `S` be the sample space.
Then, `n(S) = ` Total number of ways of drawing `4` balls out of `15`
`:.n(S) = {::}^15C_4=(15*14*13*121213*121214*13*121213*1212*11!)/((4*3*2*1)*(11!))=(15*14*13*121213*121214*13*121213*1212)/(4*3*2*1)=1365`
Let `E=` Event of drawing 2 black balls out of 4 and 2 other balls out of 11.
`n(E) = {::}^4 C_2 xx {::}^11 C_2`
`=(4*33*2!)/((2*1)*(2!)) xx (11*1010*9!)/((2*1)*(9!))`
`=6 xx 55`
`=330`
`:.P(E)=(n(E))/(n(S))=330/1365=22/91`
This material is intended as a summary. Use your textbook for detail explanation.
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