Three numbers are chosen from 1 to 8. Find the probability for the 3 chosen numbers to be consecutive. |
A) \[\frac{11}{28}\]
B) \[\frac{3}{28}\]
C) \[\frac{5}{28}\]
D) \[\frac{9}{28}\]
Correct Answer: B
Solution :
The total number of ways of choosing 3 number out of \[8={}^{8}{{C}_{3}}\]way |
Let E be the event of the three numbers to b consecutive |
\[E=\{(1,2,3),\]\[(2,3,4),\]\[(3,4,5),\]\[(4,5,6),\]\[(5,6,7),\]\[(6,7,8)\}\] |
\[\therefore \]\[P\,\,(E)=\frac{n\,\,(E)}{n\,\,(S)}=\frac{6}{{}^{8}{{C}_{3}}}=\frac{6}{56}=\frac{3}{28}\] |
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