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#### Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.  # A fair coin is to be tossed twice and an integer is to be

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Intern  Joined: 18 Mar 2012
Posts: 45
GPA: 3.7
A fair coin is to be tossed twice and an integer is to be  [#permalink]

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3
6 00:00

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
Math Expert V
Joined: 02 Sep 2009
Posts: 62289
Re: A fair coin is to be tossed twice and an integer is to be  [#permalink]

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3
4
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.

Hope it's clear.
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Manager  Joined: 24 Sep 2012
Posts: 80
Location: United States
GMAT 1: 730 Q50 V39
GPA: 3.2
WE: Education (Education)
Re: A fair coin is to be tossed twice and an integer is to be  [#permalink]

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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
Manager  Joined: 17 Apr 2013
Posts: 56
Location: United States
Concentration: Other, Finance
Schools: SDSU '16
GMAT 1: 660 Q47 V34
GPA: 2.76
WE: Analyst (Real Estate)
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

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
did get the result by luck ? Thx Math Expert V
Joined: 02 Sep 2009
Posts: 62289
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
did get the result by luck ? 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.

Overall = 2/6 + 1/6.
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Manager  Joined: 11 Jun 2014
Posts: 52
Concentration: Technology, Marketing
GMAT 1: 770 Q50 V45
WE: Information Technology (Consulting)
Re: A fair coin is to be tossed twice and an integer is to be  [#permalink]

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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
Senior Manager  Joined: 20 Aug 2015
Posts: 379
Location: India
GMAT 1: 760 Q50 V44
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
Manager  G
Joined: 14 Oct 2012
Posts: 157
A fair coin is to be tossed twice and an integer is to be  [#permalink]

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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... 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...
Non-Human User Joined: 09 Sep 2013
Posts: 14448
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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