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A fair coin is to be tossed twice and an integer is to be

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A fair coin is to be tossed twice and an integer is to be  [#permalink]

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New post 05 Mar 2013, 13:11
3
6
00:00
A
B
C
D
E

Difficulty:

  55% (hard)

Question Stats:

57% (01:27) correct 43% (01:30) wrong based on 299 sessions

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

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New post 06 Mar 2013, 01:23
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alexpavlos 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


The probability that we have at least one head in two flips = 1- P(no head in two flips) = 1- 1/2*1/2 = 3/4.

The probability of selecting even number from {3, 8, 10} = 2/3.

The probability both event 1 AND even t 2 occur is therefore 3/4*2/3=1/2.

Answer: D.

Hope it's clear.
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Re: A fair coin is to be tossed twice and an integer is to be  [#permalink]

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New post 05 Mar 2013, 14:55
Hello Alexpavlos,

Let me try helping you with this one. This question can be solved in quite a few ways.
1) Let us simply state the possibilities. This method works for this problem due to the limited number of possibilities

Ways of getting at least a head and an even number =6
HT8
TH8
HT10
TH10
HH8
HH10

Ways of not getting at least one head and an even number=6
TT8
TT10
HT3
TH3
HH3
TT3

Total probability=6/12=1/2

Method 2)
Probability of getting at least one head and an even number= probability of getting head*probability of getting tail *probability of getting an even number+probability of getting tail*probability of getting head*probability of getting an even number+probability of getting head*probability of getting head*probability of getting an even number
=(1/2)*(1/2)*(2/3)+(1/2)*(1/2)*(2/3)+(1/2)*(1/2)*(2/3)=1/2

Hope this helps! Let me know if I can help you further.


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

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New post 25 Jun 2014, 21:40
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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New post 26 Jun 2014, 00:48
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]

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New post 26 Jun 2014, 07:20
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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New post 16 Nov 2015, 02:10
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]

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New post 04 Sep 2016, 09:28
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]

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Re: A fair coin is to be tossed twice and an integer is to be   [#permalink] 28 Oct 2019, 19:32
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