A --> 3/4, which is Greater than B


So A is greater.

May you pls see my question and answers where is my fall exactly ?

QUANTITY A: The probability that the coin will NOT land on heads both times

There are 3 ways in which the coin does NOT land on heads both times
1) Heads on 1st toss AND tails on 2nd toss (here, we do NOT have heads on BOTH tosses)
2) Tails on 1st toss AND heads on 2nd toss (here, we do NOT have heads on BOTH tosses)
3) Tails on 1st toss AND tails on 2nd toss (here, we do NOT have heads on BOTH tosses)

Given that A fair coin will be tossed twice.

Quantity A : The probability that the coin will NOT land on heads both times

Probability that coin will not land into head twice = 1 - Probability(Coin will land up into heads twice) = 1 - P(2H)

Now, there are 4 case TT, TH, HT, HH

=> Probability that coin will not land into head twice = 1 - \(\frac{1}{4}\) = \(\frac{3}{4}\) = \(\frac{6}{8}\)

Clearly, Quantity A(\(\frac{6}{8}\)) > Quantity B(\(\frac{5}{8}\))

So, Answer will be A
Hope it helps!

Watch the following video to learn How to Solve Probability with Coin Toss Problems

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