Will be gratefull for help in solving 2 questions:
1. When a coin is tossed for 6 times, what is the probability that after the first tossing, every outcome will be different from previous one?
The first toss can be anything H or T. Probability = 1
The second toss must be the opposite. Say if first was H, the second must be T. Probability = 1/2
The third must be opposite again i.e. H now. Probability = 1/2
The fourth must be T. Probability = 1/2
Fifth must be H. Probability = 1/2
Sixth must be T. Probability = 1/2
So overall probability = \(1*(1/2)*(1/2)*(1/2)*(1/2)*(1/2) = 1/2^5\)
Or think of it this way.
The favorable outcomes are only two: HTHTHT or THTHTH
The total outcomes = \(2^6\)
Probability \(= 2/2^6 = 1/2^5\)
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