A) \[\frac{1}{2}\]
B) \[\frac{49}{101}\]
C) \[\frac{50}{101}\]
D) \[\frac{51}{101}\]
Correct Answer: D
Solution :
[d] Let \[X\tilde{\ }B(100,p)\] be the number of coins showing heads, and let \[q=1-p.\]Then, since \[P(X=51)=P(X=50),\]we have \[^{100}{{C}_{51}}({{P}^{51}})({{q}^{49}}){{=}^{100}}{{C}_{50}}({{p}^{50}})({{q}^{50}})\] \[\Rightarrow \frac{p}{q}=\left( \frac{100!}{50!50!} \right)\left( \frac{51!49!}{100!} \right)\] \[\Rightarrow \frac{p}{1-p}=\frac{51}{50}\Rightarrow 50p=51-51p\Rightarrow p=\frac{51}{101}\]
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