Five coins whose faces are marked 2, 3 are tossed. The chance of obtaining a total of 12 is |
A) \[\frac{1}{32}\]
B) \[\frac{1}{16}\]
C) \[\frac{3}{16}\]
D) \[\frac{5}{16}\]
Correct Answer: C
Solution :
The probability of getting a number either 2 or 3 in one toss is \[\frac{1}{2}.\] |
Condition for getting the sum of 12 in five tosses is (2, 2, 2, 3, 3). |
\[\therefore \] Required probability \[={}^{5}{{C}_{3}}{{\left( \frac{1}{2} \right)}^{3}}{{\left( \frac{1}{2} \right)}^{2}}\] |
\[=\,\,\frac{5\times 4}{2\times 1}\,\,{{\left( \frac{1}{2} \right)}^{5}}\] \[=10.\frac{1}{{{2}^{5}}}=\frac{5}{16}\] |
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