A) \[\frac{193}{256}\]
B) \[\frac{9}{128}\]
C) \[\frac{1}{2}\]
D) \[\frac{63}{256}\]
Correct Answer: D
Solution :
[d] Probability for a head \[=\frac{1}{2}i.e,\text{ }p=\frac{1}{2}\] \[\therefore q=\frac{1}{2}\] The probability that out of 10 coins, r coins show head \[{{=}^{10}}{{C}_{r}}{{\left( \frac{1}{2} \right)}^{r}}{{\left( \frac{1}{2} \right)}^{10-r}}\] \[\therefore \]Required probability \[{{=}^{10}}{{C}_{5}}{{\left( \frac{1}{2} \right)}^{5}}{{\left( \frac{1}{2} \right)}^{5}}\] \[=\frac{10\times 9\times 8\times 7\times 6}{1\times 2\times 3\times 4\times 5}.{{\left( \frac{1}{2} \right)}^{10}}=\frac{63}{{{2}^{8}}}=\frac{63}{256}\]
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