| A bag contains 4 white and 6 red balls. Two draws of one ball each are made without replacement. The probability that one is red and other white, is |
A) \[\frac{10}{15}\]
B) \[\frac{6}{9}\]
C) \[\frac{8}{15}\]
D) \[\frac{4}{9}\]
Correct Answer: C
Solution :
| \[{{W}_{1}}=\]Drawing white ball for first time |
| \[P\,\,({{W}_{1}})4/10\] |
| \[{{W}_{2}}=\]Drawing white ball in second time |
| \[P\,\,({{W}_{2}}/{{R}_{1}})=4/9\] |
| \[{{R}_{1}}=\]Drawing red ball for first time |
| \[P\,\,({{R}_{1}})=6/10\] |
| \[{{R}_{2}}=\]Drawing red ball for second time |
| \[P\,\,({{R}_{2}}/{{W}_{1}})=6/9\] |
| P (one red and three white) |
| \[=P\,\,({{R}_{1}})-P\,\,({{W}_{2}}/{{R}_{1}})+P\,\,({{W}_{1}})-P\,\,({{R}_{2}}/{{W}_{1}})\] |
| \[=\frac{6}{10}\times \frac{4}{9}+\frac{4}{10}\times \frac{6}{9}=\frac{48}{90}=\frac{8}{15}\] |
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