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In a bag,there are 4 fair coins and 3 unfair coins.The probability of getting a head in those unfair coins is 1/3 and tail is 2/3.Now if 2 coins are taken from the bag and flipped.What is the probability of getting both as heads ?
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  • $\begin{align*} \frac{2}{7}*\frac{1}{4} + \frac{2}{7}*\frac{1}{6} + \frac{2}{7}*\frac{1}{6} + \frac{1}{7}*\frac{1}{9} = \frac{23}{126} \\ \end{align*}$
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The two coins i choose can be 

                       A) FAIR - FAIR 

                       B) FAIR - UNFAIR 

                       C) UNFAIR - UNFAIR 

NOTE : Here order doesnot matter, since i am choosing 2 balls (fair-unfair and unfair-fair both are same)

PROBABILITY (of getting both as heads)

       = P(FF)*P(HH/FF) + P(FU)P(HH/FU) + P(UU)P(HH/UU)

       = (4/7 * 3/6 * 1/2 * 1/2) + (4/7 * 3/6 * 1/2 * 1/3) + (3/7 * 2/6 * 1/3 * 1/3)

                          

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