2. Solved Papers
  3. JAMIA MILLIA ISLAMIA

JAMIA MILLIA ISLAMIA Jamia Millia Islamia Solved Paper-2005

  • question_answer
        A biased coin with probability\[p,0<p<1\]of heads is tossed until a head appears for the first time. If the probability that the number of tosses required is even is\[\frac{2}{5},\]then p equals:

    A)  \[\frac{1}{3}\]                                  

    B)  \[\frac{2}{3}\]

    C)  \[\frac{2}{5}\]                                  

    D)  \[\frac{3}{5}\]

    Correct Answer: A

    Solution :

                    Let\[q=1-p\]. Since, head appears first time in an even throw 2 or 4 or 6. \[\therefore \]  \[\frac{2}{5}=qp+{{q}^{3}}p+{{q}^{5}}p+...\] \[\therefore \]  \[\frac{2}{5}=\frac{qp}{1-{{q}^{2}}}\] \[\Rightarrow \]               \[\frac{2}{5}=\frac{(1-p)p}{1-{{(1-p)}^{2}}}\]                 \[\frac{2}{5}=\frac{1-p}{2-p}\]                 \[\frac{2}{5}=\frac{1-p}{2-p}\] \[\Rightarrow \]               \[4-2p=5-5p\] \[\Rightarrow \]               \[3p=1\]               \[\Rightarrow \]\[p=\frac{1}{3}\]


You need to login to perform this action.
You will be redirected in 3 sec spinner