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A fair die is rolled once and a fair coin is flipped once.

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A fair die is rolled once and a fair coin is flipped once. [#permalink] New post 30 Dec 2007, 08:29
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A fair die is rolled once and a fair coin is flipped once. What is the probability that either the die will land on 3 or that the coin will land on heads?
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 [#permalink] New post 30 Dec 2007, 08:54
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There are 12 possibilities for outcomes:

Coin: H or T
Dice: 1, 2, 3, 4, 5 or 6

Odds of the coin landing on heads: 1/2
Odds of the dice landing on three when the coin lands on tails: 1/12

1/2+1/12 = 7/12 probability

Make sure you don't try to add 1/2 and 1/6. We've already counted the dice landing on 3 when the coin lands on heads (included in the 1/2), so we're looking for the chance of the coin landing on tails and the dice landing on 3, which is 1/12

H1, H2, H3, H4, H5, H6, T3 are the 7 possibilities that work
T1, T2, T4, T5, T6 are the 5 possibilities that don't work
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 [#permalink] New post 30 Dec 2007, 11:19
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I totally agree with eschn3am

and would like to add some important note that either X or Y means X alone, Y alone, and both XY (!)

P(X or Y)=P(X)+P(Y)-P(both X,Y)
where, P(X) means probability of X regardless of Y
where, P(Y) means probability of Y regardless of X

in our case: 1/2+1/6-1/12=(6+2-1)/12=7/12

there is a little bit of theory: http://richardbowles.tripod.com/maths/p ... y/prob.htm
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Re: Probability question [#permalink] New post 25 Aug 2008, 10:02
tekno9000 wrote:
A fair die is rolled once and a fair coin is flipped once. What is the probaility that either the die will land on 3 or that the coin will land on heads?

Please solve and explain.

Thanks,

Tekno9000


P = 1/2 +1/6 - (1/2)*(1/6) = 7/12
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Re: Probability question [#permalink] New post 27 Sep 2009, 10:22
A fair die is rolled once and a fair coin is flipped once. What is the probaility that either the die will land on 3 or that the coin will land on heads?

Soln: The probability that either the die will land on 3 or coin will land heads is
= Prob(die lands 3) + Prob(coin land heads) - Prob(that both happen)
= 1/6 + 1/2 - 1/2*1/6
= 7/12
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Re: Probability question [#permalink] New post 01 Feb 2010, 13:47
what about this way =

P(X or Y)= 1-P(NOT X)x P(NOTY)

1 - 5/6*1/2=7/12
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Re: A fair die is rolled once and a fair coin is flipped once. [#permalink] New post 29 Feb 2012, 10:55
I understand that the formula here is:
p(a or b) = p(a) + p(b) - p(a and b)
= 1/6 + 3/6 - (1/12)
= 4/6 - 1/12
= 7/12

But, I'm curious as to what's wrong with this logic:
p(die landing on 3) = 1/6
p(heads) = 1/2
Therefore p(die landing on 3 or coin landing on heads) = 1/6 + 1/2 = 2/3.
Why do I have to account for the time when both events occur, and subtract it from 2/3?
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Re: A fair die is rolled once and a fair coin is flipped once. [#permalink] New post 29 Feb 2012, 11:48
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fortsill wrote:
I understand that the formula here is:
p(a or b) = p(a) + p(b) - p(a and b)
= 1/6 + 3/6 - (1/12)
= 4/6 - 1/12
= 7/12

But, I'm curious as to what's wrong with this logic:
p(die landing on 3) = 1/6
p(heads) = 1/2
Therefore p(die landing on 3 or coin landing on heads) = 1/6 + 1/2 = 2/3.
Why do I have to account for the time when both events occur, and subtract it from 2/3?


Because 1/6 is basically 1/6*1: the die landing on 3 and the coin landing on any side. The same way 1/2 is basically 1/2*1: the coin landing on heads and the die landing on any side. So, both 1/6 and 1/2 count (include) the probability of the case when the die lands on 3 and the coin lands on heads, hence we should subtract P(3 on the die and heads on the coin)=1/6*1/2 once, to avoid double counting.

P=1/6+1/2-1/6*1/2=7/12.

A fair die is rolled once and a fair coin is flipped once. What is the probaility that either the die will land on 3 or that the coin will land on heads?

This question can be solved with an easier approach: P(3 on a die OR heads on a coin)=1-P(neither 3 on a die nor heads on a coin)=1-5/6*1/2=7/12.

OR probability:
If Events A and B are independent, the probability that Event A OR Event B occurs is equal to the probability that Event A occurs plus the probability that Event B occurs minus the probability that both Events A and B occur: P(A \ or \ B) = P(A) + P(B) - P(A \ and \ B).

This is basically the same as 2 overlapping sets formula:
{total # of items in groups A or B} = {# of items in group A} + {# of items in group B} - {# of items in A and B}.

Note that if event are mutually exclusive then P(A \ and \ B)=0 and the formula simplifies to: P(A \ or \ B) = P(A) + P(B).

Also note that when we say "A or B occurs" we include three possibilities:
A occurs and B does not occur;
B occurs and A does not occur;
Both A and B occur.

AND probability:
When two events are independent, the probability of both occurring is the product of the probabilities of the individual events: P(A \ and \ B) = P(A)*P(B).

This is basically the same as Principle of Multiplication: if one event can occur in m ways and a second can occur independently of the first in n ways, then the two events can occur in mn ways.

Hope it helps.
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Re: A fair die is rolled once and a fair coin is flipped once. [#permalink] New post 14 Apr 2012, 13:55
The above formula is incorrect. Let me explain and then prove it. When you add the probabilities for the dice hitting 3 (1/6) to the coin toss resulting in a heads (1/2) and then subtract the probability of BOTH a heads and a 3 (1/12), you aren't factoring that that probability was counted TWICE in your original sum. Once during your probability for 3 and once during your probability for a heads result.

Here is the proof.. there are 12 possibilities. I've put an asterisk next to the ones which satisfy the scenario (#s signify dice and t is tails, h is heads):

1t
1h*
2t
2h*
3t*
3h
4t
4h*
5t
5h*
6t
6h*

3h is obviously excluded because it includes both heads and 3. Now if you see, there are only 12 possible scenarios, and only 6 satisfy our condition (heads OR 3). Therefore, the probability is 6/12 = 1/2.

Now, I've thought of a simpler way of approaching these type of problems...
Probability that it will be a HEADS and NOT A 3...
1/2 x 5/6 = 5/12

Probability that it will be a 3 and NOT HEADS...
1/6 x 1/2 = 1/12

ADD probabilities...
1/12 + 5/12 = 6/12 = 1/2
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Re: A fair die is rolled once and a fair coin is flipped once. [#permalink] New post 14 Apr 2012, 14:05
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Germainte wrote:
The above formula is incorrect. Let me explain and then prove it. When you add the probabilities for the dice hitting 3 (1/6) to the coin toss resulting in a heads (1/2) and then subtract the probability of BOTH a heads and a 3 (1/12), you aren't factoring that that probability was counted TWICE in your original sum. Once during your probability for 3 and once during your probability for a heads result.

Here is the proof.. there are 12 possibilities. I've put an asterisk next to the ones which satisfy the scenario (#s signify dice and t is tails, h is heads):

1t
1h*
2t
2h*
3t*
3h
4t
4h*
5t
5h*
6t
6h*

3h is obviously excluded because it includes both heads and 3. Now if you see, there are only 12 possible scenarios, and only 6 satisfy our condition (heads OR 3). Therefore, the probability is 6/12 = 1/2.

Now, I've thought of a simpler way of approaching these type of problems...
Probability that it will be a HEADS and NOT A 3...
1/2 x 5/6 = 5/12

Probability that it will be a 3 and NOT HEADS...
1/6 x 1/2 = 1/12

ADD probabilities...
1/12 + 5/12 = 6/12 = 1/2


This is MGMAT question (from Manhattan GMAT Strategy Guide) and the OA is 7/12, not 1/2. I guess your reading of the question in different from that of the autors of the question.
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Re: A fair die is rolled once and a fair coin is flipped once. [#permalink] New post 15 Apr 2012, 10:16
Rolling a dice and throwing a coin are independent events, because they can occur at the same time, right? In the probability part of the GMAT Club mathbook it says that two events are independent if the occurence of one event does not influence the occurence of the other event. Then it says that "tossing a coin and rolling a die are independent events". So what's correct? And if they were independent, we wouldn't have to subtract the probability of both occuring at same time, would we?
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Re: A fair die is rolled once and a fair coin is flipped once. [#permalink] New post 15 Apr 2012, 10:18
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BN1989 wrote:
Rolling a dice and throwing a coin are independent events, because they can occur at the same time, right? In the probability part of the GMAT Club mathbook it says that two events are independent if the occurence of one event does not influence the occurence of the other event. Then it says that "tossing a coin and rolling a die are independent events". So what's correct? And if they were independent, we wouldn't have to subtract the probability of both occuring at same time, would we?


Explained here: a-fair-die-is-rolled-once-and-a-fair-coin-is-flipped-once-57799.html#p1051919
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Re: A fair die is rolled once and a fair coin is flipped once. [#permalink] New post 17 Apr 2012, 05:45
BN1989 wrote:
Rolling a dice and throwing a coin are independent events, because they can occur at the same time, right? In the probability part of the GMAT Club mathbook it says that two events are independent if the occurence of one event does not influence the occurence of the other event. Then it says that "tossing a coin and rolling a die are independent events". So what's correct? And if they were independent, we wouldn't have to subtract the probability of both occuring at same time, would we?


See it this way, both the events can happen together. and that prob would be 1/2*1/6 . What you mentioned is the case when both cannot occur together . e.g. prob of winning a race for a or b = prob(a) + prob(b)


Two events, A and B, are independent if the fact that A occurs does not affect the probability of B occurring. BUT they can occur together and i.e. prob of occurrence together is prob(a)*prob(b) ( product of their individual prob)
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Re: A fair die is rolled once and a fair coin is flipped once. [#permalink] New post 31 Jul 2013, 04:34
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Thumb rule :
1. Events which can occur in one attempt subtract their double counting.
E.g : In above case dice and coin can be flipped together and we can get three possible outcomes.
1 head no 3
1 tail and 3
1 head and 3 (thus both are overlapping and its possible in one attempt)
E.g : Choose Spade or a king, overlapping is at King of spade.

2. Events which can not occur in one attempt, simply add their probabilities.
E.g : Draw at least 2 red balls in 4 draws :
case 1 : 2 red ball 2 black ball.
case 2 : 3 red ball 1 black ball.
case 4 : 4 red ball 0 black ball.
These all cases can not occur in one attempt, thus simply add their probability, no overlapping.

I tried to understand this concept in this way.
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Re: A fair die is rolled once and a fair coin is flipped once. [#permalink] New post 07 Jan 2014, 15:27
How about this approach?

There are 3 desired scenarios:

scenario 1: Die lands on 3 AND coin DOESN'T land on tails.
scenario 2: Die DOESN'T land on 3 AND coin lands on heads.
scenario 3: Die lands on 3 AND coin lands on heads.

So I add the probability of the 3 scenarios:

(scenario 1) + (scenario 2) + (scenario 3)

(Die condition * coin condition) + (Die condition * coin condition) + (Die condition * coin condition)

(1/6 * 1/2) + (5/6 * 1/2) + (1/6 * 1/2) = 7/12

scenario 1= (1/6 * 1/2) Only 3 is the desired outcome AND only tails is the desired outcome (To meet the condition of NO heads).
scenario 2= (5/6 * 1/2) All number are desired outcomes but 3 (To meet the condition of NO 3) AND only heads is the desired outcome.
scenario 3= (1/6 * 1/2) Only 3 is the desired outcome AND only heads is the desired outcome.

At the end of the day is the same as applying the formula. But is easier for me to "prevent" the overlapping count and forget about the last step of subtracting this overlap.
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Re: A fair die is rolled once and a fair coin is flipped once. [#permalink] New post 07 Jan 2014, 19:25
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How about this approach?

There are 3 desired scenarios:

scenario 1: Die lands on 3 AND coin DOESN'T land on tails.
scenario 2: Die DOESN'T land on 3 AND coin lands on heads.
scenario 3: Die lands on 3 AND coin lands on heads.

So I add the probability of the 3 scenarios:

(scenario 1) + (scenario 2) + (scenario 3)

(Die condition * coin condition) + (Die condition * coin condition) + (Die condition * coin condition)

(1/6 * 1/2) + (5/6 * 1/2) + (1/6 * 1/2) = 7/12

scenario 1= (1/6 * 1/2) Only 3 is the desired outcome AND only tails is the desired outcome (To meet the condition of NO heads).
scenario 2= (5/6 * 1/2) All number are desired outcomes but 3 (To meet the condition of NO 3) AND only heads is the desired outcome.
scenario 3= (1/6 * 1/2) Only 3 is the desired outcome AND only heads is the desired outcome.

At the end of the day is the same as applying the formula. But is easier for me to "prevent" the overlapping count and forget about the last step of subtracting this overlap.


This approach is fine too but here is a problem with enumerating cases - we might forget a case or two under time pressure.
This question is extremely simple and you could think of overlapping sets here. The question asks you the probability of A or B or both. This is your case of the number of elements which are in set A or set B or both. How do you calculate that?
n(A) + n(B) - n(A and B) (because n(A and B) is counted twice - once in n(A) and another time in n(B) so you need to subtract it out once)
This is the same concept.
P(A) + P(B) - P(A and B) = 1/6 + 1/2 - (1/6*1/2) = 7/12
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Re: A fair die is rolled once and a fair coin is flipped once.   [#permalink] 07 Jan 2014, 19:25
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