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

Question

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

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

Bunuel · Accepted Answer

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.

Kris01 · Answer

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.

clipea12 · Answer

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

Bunuel · Answer

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.

sshrivats · Answer

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

TeamGMATIFY · Answer

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

manishtank1988 · Answer

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...

P(2T & odd) = P(2T)*P(odd) = (1/2*1/2)*(1/3) but, P = = = = 1/2 If you like it, kudo...

CEdward · Answer

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.

N0BU · Answer

Bunuel   Krunaal   alex1233  not sure if the "Inequalities" got accidently added seems like it just needs the probability tag.

Bunuel · Answer

Removed the tag. Thank you!

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

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