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Intern  B
Joined: 15 Mar 2010
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Two dice are tossed once. The probability of getting an even number at  [#permalink]

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10
63 00:00

Difficulty:   95% (hard)

Question Stats: 42% (02:14) correct 58% (02:14) wrong based on 910 sessions

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Two dice are tossed once. The probability of getting an even number at the first die or a total of 8 is

A. 1/36
B. 3/36
C. 11/36
D. 20/36
E. 23/36

Originally posted by Macsen on 23 Jan 2012, 21:20.
Last edited by Bunuel on 04 Feb 2019, 05:13, edited 1 time in total.
Renamed the topic.
Math Expert V
Joined: 02 Sep 2009
Posts: 58344
Re: Two dice are tossed once. The probability of getting an even number at  [#permalink]

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9
15
Macsen wrote:
Two dice are tossed once. The probability of getting an even number at the first die or a total of 8 is
A. 1/36
B. 3/36
C. 11/36
D. 20/36
E. 23/36

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.

BACK TO THE ORIGINAL QUESTION:
Two dice are tossed once. The probability of getting an even number at the first die or a total of 8 is
A. 1/36
B. 3/36
C. 11/36
D. 20/36
E. 23/36

1. The probability of getting an even number at the first die is 1/2 (as the probability of even = the probability of odd = 1/2);
2. The probability of getting a total of 8 is 5/6^2, as there are 5 different favorable scenarios: (2,6), (6,2), (3,5), (5,3) and (4,4);
3. The probability of getting an even number at the first die AND a total of 8 is 3/6^2 (from above case);

Hence, The probability of getting an even number at the first die OR a total of 8 is 1/2+5/36-3/36=20/36.

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Posts: 6
Re: Two dice are tossed once. The probability of getting an even number at  [#permalink]

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6
5
total possible outcomes for two dice: 36

case 1
prob of getting an even number on the first die
first die can have 2,4,6 second die can have any of 1,2,3,4,5,6
or favorable outcomes 3x6 = 18
probability= 18/36

case 2
getting a sum of 8 (2,6)(6,2)(3,5)(5,3)(4,4)
probability=5/36

now both of the above cases have some cases common to them
i.e. when the first die has an even number and the sum is also 8
there are 3 cases of this kind (2,6) (6,2) (4,4)
prob=3/36

also P(A or B)=P(A) + P(B) - P(A & B)
so we have P(even or sum of 8) = 18/36 + 5/36 - 3/36
20/36
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Re: Two dice are tossed once. The probability of getting an even number at  [#permalink]

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+1 D

Great question.
Remember that the "OR formula" is:

P(A) OR P(B) = P(A) + P(B) - P(A and B)

In this case, A and B can take place together, so the value of P(A and B) is greater than 0.
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Re: Two dice are tossed once. The probability of getting an even number at  [#permalink]

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Bunuel, could you please explain why the probability of getting 'four' on both dices is 1/36 and not 1/18? Why don't we make a difference between 4 on dice 1 and four on dice ?

I mean why the combination 4'4'' is the same as 4''4' (apostrophe symbolizes the dice: ' = dice number one, '' = dice number two).

I have encountered many probability problems, where events that are identical are differentiated. For example:

Quote:
A 5 meter long wire is cut into two pieces. The longer piece is then used to form a perimeter of a square. What is the probability that the area of the square will be more than 1 if the original wire is cut at an arbitrary point?

1/6
1/5
3/10
1/3
2/5

OA:
(E)

In the example above the probability is 2/5.

Could you please explain the difference?

Originally posted by nonameee on 21 Feb 2012, 02:10.
Last edited by nonameee on 21 Feb 2012, 02:24, edited 1 time in total.
Math Expert V
Joined: 02 Sep 2009
Posts: 58344
Re: Two dice are tossed once. The probability of getting an even number at  [#permalink]

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nonameee wrote:
Bunuel, could you please explain why the probability of getting 'four' on both dices is 1/36 and not 1/18? Why don't we make a difference between 4 on dice 1 and four on dice ?

I mean why the combination 4'4'' is the same as 4''4' (apostrophe symbolizes the dice: ' = dice number one, '' = dice number two).

We can get combination of (2, 6) in two ways: 2 on die A and 6 on die B OR 2 on die B and 6 on die A;

Whereas (4, 4) has only one combination: 4 on on die A and 4 on die B, it has no second combination, since 4 on die B and 4 on die A is exact same combination.

nonameee wrote:
I have encountered many probability problems, where events that are identical are differentiated. For example:

Quote:
A 5 meter long wire is cut into two pieces. The longer piece is then used to form a perimeter of a square. What is the probability that the area of the square will be more than 1 if the original wire is cut at an arbitrary point?

1/6
1/5
3/10
1/3
2/5

OA:
(E)

In the example above the probability is 2/5.

Could you please explain the difference?

This is completely different problem. In order the area of a square to be more than 1 the side of it must be more than 1, or the perimeter more than 4. So the longer piece must be more than 4. Look at the diagram.

-----

If the wire will be cut anywhere at the red region then the rest of the wire (longer piece) will be more than 4 meter long. The probability of that is 2/5 (2 red pieces out of 5).

Also discussed here: a-5-meter-long-wire-is-cut-into-two-pieces-if-the-longer-106448.html

Hope it helps.
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Posts: 58344
Re: Two dice are tossed once. The probability of getting an even number at  [#permalink]

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Re: Two dice are tossed once. The probability of getting an even number at  [#permalink]

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I've done other problems in which the answer would simply come out from

1/2 + (1/2)(5/36)

That is first probability that we get an even number = 1/2
Second probability that we DON'T get an even number * Probability that we get a sum of 8 = (1/2)*(5/36)

Answer should be sum of both

Could someone please clarify why this approach is NOT valid?
Thanks!
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Joined: 30 Jul 2014
Posts: 107
GPA: 3.72
Re: Two dice are tossed once. The probability of getting an even number at  [#permalink]

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It is a very good question - I forgot to subtract the 3 cases in which the second dice will also show an even number - hence marked E, when the answer should be D...
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Re: Two dice are tossed once. The probability of getting an even number at  [#permalink]

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no calc required

half of the time the first dice will have even no. so prob will be greater than 1/2 as some cases extra when sum is 8.
so A,B,C ruled out

E gives 5 cases for sum 8.
but we know that sum cases will be common with "even condition" so D
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Re: Two dice are tossed once. The probability of getting an even number at  [#permalink]

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Macsen wrote:
Two dice are tossed once. The probability of getting an even number at the first die or a total of 8 is

A. 1/36
B. 3/36
C. 11/36
D. 20/36
E. 23/36

When we roll two dice, there are 6 x 6 = 36 equally likely outcomes. From these 36 outcomes, we can get a total of 8 in the following ways:

6,2

2,6

4,4

5,3

3,5

The probability of each of these outcomes is 1/36, so the probability of getting a total of 8 is 5/36.

The probability of getting an even number on the first die is 1/2.

Recall that the probability of A or B is P(A or B) = P(A) + P(B) - P(A and B).

Let’s concentrate on P(A and B), which includes outcomes that have an even first roll AND a total of 8. The probability that the first die is even and the total of the two dice is 8 is 3/36 since (6,2), (2,6), and (4,4) are 3 of the 36 outcomes in which the first die is even and the total is 8.

Thus, the total probability of getting an even number on the first die OR a total of 8 is:

5/36 + 1/2 - 3/36 = 5/36 + 18/36 - 3/36 = 20/36

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Joined: 08 Dec 2017
Posts: 15
Re: Two dice are tossed once. The probability of getting an even number at  [#permalink]

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We van just list out the possibilites.

2,1
2,2
2,3
2,4
2,5

2,6

3,5

4,1
4,2
4,3
4,4
4,5
4,6

5,3

6,1
6,2
6,3
6,4
6,5
6,6

So 20/36. 18 cases occur from the first dice being even. The other two cases are 3,5 and 5,3

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Re: Two dice are tossed once. The probability of getting an even number at  [#permalink]

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_________________ Re: Two dice are tossed once. The probability of getting an even number at   [#permalink] 04 Feb 2019, 05:15
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