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Re: A fair coin will be tossed twice [#permalink]
A --> 3/4, which is Greater than B


So A is greater.
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Re: A fair coin will be tossed twice [#permalink]
GreenlightTestPrep wrote:
Asmakan wrote:
A fair coin will be tossed twice.
Quantity A
Quantity B
The probability that the coin will NOT land on heads both times
\(\frac{5}{8}\)


We can use the complement here.
That is, P(Event A happening) = 1 - P(Event A not happening)
We can rewrite this as: P(Event A not happening) = 1 - P(Event A happening)

So, P(NOT getting two heads) = 1 - P(getting two heads)

P(getting two heads)
P(getting two heads) = P(heads on 1st toss AND heads on 2nd toss)
= P(heads on 1st toss) x P(heads on 2nd toss)
= 1/2 x 1/2
= 1/4

So, P(NOT getting two heads) = 1 - 1/4
= 3/4

We get:
QUANTITY A: 3/4 (aka 6/8)
QUANTITY B: 5/8

Answer: A

Cheers,
Brent



May you pls see my question and answers where is my fall exactly ?
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Re: A fair coin will be tossed twice [#permalink]
2
Asmakan wrote:
I have solved the question by this way :
Quantity A means that the coin will be tails on both sides. This means ( 0.5 x 0.5)= 0.25

May someone explains why did I got it wrong and the steps to solve it


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)
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Re: A fair coin will be tossed twice [#permalink]
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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Re: A fair coin will be tossed twice [#permalink]
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