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# A fair die is rolled once and a fair coin is flipped once. What is the

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Re: A fair die is rolled once and a fair coin is flipped once. What is the [#permalink]
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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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Re: A fair die is rolled once and a fair coin is flipped once. What is the [#permalink]
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tekno9000
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: A fair die is rolled once and a fair coin is flipped once. What is the [#permalink]
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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: A fair die is rolled once and a fair coin is flipped once. What is the [#permalink]
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. What is the [#permalink]
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. What is the [#permalink]
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. What is the [#permalink]
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Germainte
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. What is the [#permalink]
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. What is the [#permalink]
Expert Reply
BN1989
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. What is the [#permalink]
BN1989
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. What is the [#permalink]
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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. What is the [#permalink]
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. What is the [#permalink]
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Carmona
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. What is the [#permalink]
I still quite don't understand this, my logic is this:

P that BOTH Events occur: 1/6 * 1/2 = 1/12
P that NONE of the Events occur: 5/6 * 1/2 = 5/12 ;that means no 3 and no heads

So 1 - 6/12 = 6/12 which is that either one or the other event must occur. 3/noH or no3/H
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Re: A fair die is rolled once and a fair coin is flipped once. What is the [#permalink]
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madmax1000
I still quite don't understand this, my logic is this:

P that BOTH Events occur: 1/6 * 1/2 = 1/12
P that NONE of the Events occur: 5/6 * 1/2 = 5/12 ;that means no 3 and no heads

So 1 - 6/12 = 6/12 which is that either one or the other event must occur. 3/noH or no3/H

"A occurs or B occurs" implies that either only A occurs or only B occurs or both occur. In logic, "or" includes "both". Hence you will not subtract 6/12 but only 5/12 from 1 to get 1 - 5/12 = 7/12.
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Re: A fair die is rolled once and a fair coin is flipped once. What is the [#permalink]
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tekno9000
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?

Event A = Fair die is rolled once
Event B = Fair coin is flipped once.

P(A or B) = 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. What is the [#permalink]
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VeritasPrepKarishma
madmax1000
I still quite don't understand this, my logic is this:

P that BOTH Events occur: 1/6 * 1/2 = 1/12
P that NONE of the Events occur: 5/6 * 1/2 = 5/12 ;that means no 3 and no heads

So 1 - 6/12 = 6/12 which is that either one or the other event must occur. 3/noH or no3/H

"A occurs or B occurs" implies that either only A occurs or only B occurs or both occur. In logic, "or" includes "both". Hence you will not subtract 6/12 but only 5/12 from 1 to get 1 - 5/12 = 7/12.

hi,
either A or B should not involve a scenario where both occur....
incase both together is also a scenario, the question should have been..."ways in which atleast one of the events occur?"...
is there any Q from OG which tests this concept so that one can read what the GMAC people expect?
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