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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

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.

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

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21 Feb 2012, 02:10

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?

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

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21 Feb 2012, 02:38

Expert's post

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 BOR2 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?

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).

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

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21 Feb 2012, 02:46

I have re-read my posts and I think it's clear now.

Quote:

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 die1 and four on die?

This is because we fix dices one and two.

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?

This is because we can cut the wire from both ends. When we fix the ends, we can cut 1 m from the left, or we can cut 1 m from the right end. Both options are different since we have fixed the ends.

Quote:

When calculating probability, how should we know when to count the same instances and when not?

We don't count twice the same instances of the fixed "property" (don't know how to call that). But if two fixed properties (e.g., two dices: die_1 = property_one, die_2 = property_two) yield the same result by taking various values, we need to count those options (as in the above example with dices producing the sum of 8).

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

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21 Feb 2012, 04:20

Expert's post

nonameee wrote:

I have re-read my posts and I think it's clear now.

Quote:

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 die1 and four on die?

This is because we fix dices one and two.

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?

This is because we can cut the wire from both ends. When we fix the ends, we can cut 1 m from the left, or we can cut 1 m from the right end. Both options are different since we have fixed the ends.

Quote:

When calculating probability, how should we know when to count the same instances and when not?

We don't count twice the same instances of the fixed "property" (don't know how to call that). But if two fixed properties (e.g., two dices: die_1 = property_one, die_2 = property_two) yield the same result by taking various values, we need to count those options (as in the above example with dices producing the sum of 8).

Bunuel, does the above make sense?

I'm not sure that I understand completely the last part of your post, though it seems that you got the main point. _________________

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);

Dear Bunuel, my question pertains to the quoted text.

My reasoning to get sum 8 is: Select any 5 numbers from first dice ie. probability: 5/6 To make the sum 8, only a unique number is to be chosen from 2nd dice. Hence the probability of selecting 1 number in second dice is : 1/6 Probability for sum 8 is 5/6^2. However I multiplied this number by 2 since you can perform the same task by selecting the second dice first and choosing a corresponding unique value in 1st dice. Net probability is 5*2/6^2, but i believe this is wrong. Please let me know under what situations do we employ my understanding. Sorry if it is a silly question. Thank you for your time.

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

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23 Apr 2014, 06:15

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! Cheers J

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