Question

what is the total result of bouncing two coins twise.​

Answer 1

There are 2^4 = 16 possible outcomes from 2 fair coins, fairly tossed twice and there are 2^2=4 possible outcomes from the same 2 coins tossed once.

With regard to the 1st toss, the chance of 0H = [2!/(2!)(0!)]/4 = 1/4. The chance of 1H = [2!/(1!)(1!)]/4 = 1/2. The chance of 2H = [2!/(0!)(2!)]/4 = 1/4. Therefore, the chance of just 1H = 1/2 and the chance of at least 1H = (1/2)+(1/4) = 3/4, depending on the intent of the question.

With regard to the 2nd toss, the possible outcomes and probabilities are just the same as for the 1st. Linking both together though requires manipulation of probabilities to account for probabilities of same faces occurring after 2nd toss. Chance of another 0H after 2nd toss = (1/4)*(1/4) = 1/16. Chance of another 1H after 2nd toss = (1/2)*(1/2) = 1/4. Chance of another 2H after 2nd toss = (1/4)*(1/4) = 1/16.

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