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Two dice are tossed once. The probability of getting an even number at
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Updated on: 04 Feb 2019, 04:13
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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
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Originally posted by Macsen on 23 Jan 2012, 20:20.
Last edited by Bunuel on 04 Feb 2019, 04:13, edited 1 time in total.
Renamed the topic.




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Re: Two dice are tossed once. The probability of getting an even number at
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24 Jan 2012, 01:47
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 isA. 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+5/363/36=20/36. Answer: D.
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Re: Two dice are tossed once. The probability of getting an even number at
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23 Jan 2012, 20:58
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
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19 Feb 2012, 07:18
+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
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Updated on: 21 Feb 2012, 01:24
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: In the example above the probability is 2/5. Could you please explain the difference?
Originally posted by nonameee on 21 Feb 2012, 01:10.
Last edited by nonameee on 21 Feb 2012, 01:24, edited 1 time in total.



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Re: Two dice are tossed once. The probability of getting an even number at
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21 Feb 2012, 01:38
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: 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). Answer: E. Also discussed here: a5meterlongwireiscutintotwopiecesifthelonger106448.htmlHope it helps.
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Re: Two dice are tossed once. The probability of getting an even number at
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07 Jun 2013, 06:03
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Re: Two dice are tossed once. The probability of getting an even number at
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23 Apr 2014, 05:15
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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Re: Two dice are tossed once. The probability of getting an even number at
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07 Sep 2017, 04:42
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
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08 Sep 2017, 03:20
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
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11 Sep 2017, 10:35
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 Answer: D
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Re: Two dice are tossed once. The probability of getting an even number at
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27 Dec 2017, 15:51
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
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08 Feb 2020, 23:28
Bunuel wrote: 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 isA. 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+5/363/36=20/36. Answer: D. Hi Bunuel, I understood your explanation, but I am confused with the application of the formula i.e. for two Independent events P(A and B) = P(A) x P(B). If we apply the formula in the above example, we get 1/2 x 5/36 = 5/72 and not 3/36. Can you please explain if I am missing something?




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