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Re: A fair coin is to be tossed twice and an integer is to be [#permalink]
alex1233 wrote:
A fair coin is to be tossed twice and an integer is to be selected at random from one of the integers 3, 8, and 10. What is the probability that at least one head is tossed and an even integer is selected?

A. 1/6
B. 1/4
C. 1/3
D. 1/2
E. 7/12


Hi all,

I'm not quiet sure whether my reasoning is correct or not
I assumed two cases :
1) 1 head + 1 even = 1/2 * 2/3 = 2/6
2) 2 heads + 1 even = 1/2 * 1/2 * 2/3 = 1/6
adding both the scenarios will give us 1/2 (answer D )
did get the result by luck ? :roll:

Thx :)
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Re: A fair coin is to be tossed twice and an integer is to be [#permalink]
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clipea12 wrote:
alex1233 wrote:
A fair coin is to be tossed twice and an integer is to be selected at random from one of the integers 3, 8, and 10. What is the probability that at least one head is tossed and an even integer is selected?

A. 1/6
B. 1/4
C. 1/3
D. 1/2
E. 7/12


Hi all,

I'm not quiet sure whether my reasoning is correct or not
I assumed two cases :
1) 1 head + 1 even = 1/2 * 2/3 = 2/6
2) 2 heads + 1 even = 1/2 * 1/2 * 2/3 = 1/6
adding both the scenarios will give us 1/2 (answer D )
did get the result by luck ? :roll:

Thx :)


We are told that a coin is to be tossed twice. So, the case with one head means 1 head and 1 tail. So, if you do this way it should be:

1. P(head, tail) * P(even) = (2*1/2*1/2)*2/3 = 2/6. We should multiply 1/2*1/2 by 2, because HT can occur in two ways HT and TH.

2. P(head, head) * P(even) = (1/2*1/2)*2/3 = 1/6.

Overall = 2/6 + 1/6.
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Re: A fair coin is to be tossed twice and an integer is to be [#permalink]
This is actually quite simple.

first part
a coin is tossed twice.
so probability of at least 1 head = 1 - P( all tails)
P(at least 1 head) = 1 - {(1/2)*(1/2)}
= 1 - (1/4) = 3/4


second part
there are 3 integers out of which 2 are even,
P( even integer ) = 2/3

so P( at least 1 head and 1 even integer) = (3/4) * (2/3) = 2/4 = 1/2

Ans D
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Re: A fair coin is to be tossed twice and an integer is to be [#permalink]
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alex1233 wrote:
A fair coin is to be tossed twice and an integer is to be selected at random from one of the integers 3, 8, and 10. What is the probability that at least one head is tossed and an even integer is selected?

A. 1/6
B. 1/4
C. 1/3
D. 1/2
E. 7/12


This questions can teach us two things:
1. Use the negation method in the at least questions. Subtract the probaility of undesired cases from 1 to get the probability of desired cases. It is almost always easier to solve.
2. If the question uses "AND" in between two probabilities, we need to multiply the probabilities


Probability of at least one head = 1 - P(0 Heads) = 1 - 1/4 = 3/4
We have the following cases : HH, HT, TH and TT. Only one case where there are 0 heads.

Probability of even number = 2/3

Combined probability = \frac{3}{4}* \frac{2}{3} = \frac{1}{2}
Option D
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A fair coin is to be tossed twice and an integer is to be [#permalink]
Hello Friends, I got this question correct by fluke.
What i did was this: P(2T & odd) = 1/2*1/2*1/3 = 1/12
P(atleast one T & odd) = 1-P(2T & odd) = 11/12 which isn't even the answer.
Can anyone explain me what am i doing wrong?
Can't we take both events together? If so how to do it?

Figured out what i was doing wrong: Here it goes for anyone who thought like me... :shock:
P(2T & odd) = P(2T)*P(odd) = (1/2*1/2)*(1/3)
but, P[(atleast 1 H)&(even)] = [(1- P(2 Tails) )&(1 - P(odd))] = [(1 - 1/4)*(1 - 1/3)] = [3/4 * 2/3] = 1/2

If you like it, kudo...
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Re: A fair coin is to be tossed twice and an integer is to be [#permalink]
This is how I thought about it.

a) P(1 head + 1 even) = 1/2 x 2/3 = 1/3
b) P(2 heads + 1 even) = 1/2 x 1/2 x 2/3 = 1/6

P(1 head + 1 even) + P(2 heads + 1 even) = 1/3 + 1/6 = 1/2

Answer is D.
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Re: A fair coin is to be tossed twice and an integer is to be [#permalink]
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Re: A fair coin is to be tossed twice and an integer is to be [#permalink]
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