IIT JEE , AIEEE Forum - Mathematics , Algebra

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I am assuming that coin is replaced after being tossed.

m coins are fair and N-m coins are biased with head on one side.

So probability of picking one fair coin = m/N
Probability of getting head in fair coin = (1/2)(m/N)
Probability of getting tail in fair coin = (1/2)(m/N)

Probability of picking biased coin = 1 - m/N
Probability of getting head in biased coin = (1 - m/N)
Probability of getting tail in biased coin = 0

So net probability of getting head = (1/2)(m/N) + (1 - m/N) = 1 - m/2N
and net probability of getting tail = m/2N

Question says that after 8 tosses he should be 2 steps away from the initial position.

This is possible for (5 heads + 3 tails) ; (3 heads + 5 tails).

P(5H,3T) = (1 - m/2N)⁵(m/2N)³
P(3H,5T) = (1 - m/2N)³(m/2N)⁵

so net probability = (1 - m/2N)⁵(m/2N)³ + (1 - m/2N)³(m/2N)⁵

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