How to find probability of getting two heads from a set of fair and unfair coins ?

Question

I have  4 fair and 3 unfair coins ,now I am chosing 2 coins what is the probability of getting 2 heads ,Now I am confused while calculating P(FU) i.e. probability that I chose 1st fair coin and then chose 2nd coin as unfair coin

According to me P(FU)= P(F)*P(F/U) = 4/7  *  3/ 6  since after I chose a fair coin , then will have remaining 6 choices , so what's the issue in this , then using total probability I can calculate

P(HH)=P(FF∩HH) +P(FU∩HH)+P(UU∩HH)

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

P(FF)=2/7 ,P(UU)= 1/7 other things are also stated ,only issue is with this P(FU ) , even it could be P(UF) but what's the issue since both would be same only .

Comments

For solving such questions, more information is required such as probability of getting head or tail on unfair coins.

For fair coins, prob. of getting head or tail = 1/2

but in case of unfair coin,prob. of getting head or tails should be given then only we can proceed.

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It is simply a logic of total probability

P(FU)+P(UF)=1-(P(UU)+P(FF))

Answer 1

HERE IS THE ANSWER... let Ph and Pt be the probability of unfair coin.

then 4C2/7C2[(0.5)(0.5)]+3C2/7C2[(Ph)(Ph)]+(4C1*3C1)/7C2[(0.5)*(Ph)]