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Two dice are tossed once. The probability of getting an even number at [#permalink]
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Question Stats: Hide Show timer StatisticsTwo dice are tossed once. The probability of getting an even number at the first die or a total of 8 isA. 1/36 B. 3/36C. 11/36D. 20/36 E. 23/36
Originally posted by Macsen on 23 Jan 2012, 20:20. Renamed the topic.
Math Expert Joined: 02 Sep 2009 Posts: 86746
Re: Two dice are tossed once. The probability of getting an even number at [#permalink]
Macsen wrote: 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. 3/36C. 11/36D. 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. 3/36C. 11/36D. 20/36E. 23/361. 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.Answer: D. _________________
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Re: Two dice are tossed once. The probability of getting an even number at [#permalink] total possible outcomes for two dice: 36case 1prob of getting an even number on the first diefirst die can have 2,4,6 second die can have any of 1,2,3,4,5,6 or favorable outcomes 3x6 = 18 probability= 18/36case 2getting a sum of 8 (2,6)(6,2)(3,5)(5,3)(4,4)probability=5/36now both of the above cases have some cases common to themi.e. when the first die has an even number and the sum is also 8there 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] +1 DGreat 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] 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?1/61/53/101/32/5 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.
Math Expert Joined: 02 Sep 2009 Posts: 86746
Re: Two dice are tossed once. The probability of getting an even number at [#permalink]
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?1/61/53/101/32/5 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: a-5-meter-long-wire-is-cut-into-two-pieces-if-the-longer-106448.html Hope it helps. _________________
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Re: Two dice are tossed once. The probability of getting an even number at [#permalink] Bumping for review and further discussion*. Get a kudos point for an alternative solution! *New project from GMAT Club!!! Check HERE _________________
Re: Two dice are tossed once. The probability of getting an even number at [#permalink] I've done other problems in which the answer would simply come out from1/2 + (1/2)(5/36)That is first probability that we get an even number = 1/2Second 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 bothCould someone please clarify why this approach is NOT valid?Thanks!Cheers J
Re: Two dice are tossed once. The probability of getting an even number at [#permalink] 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] no calc requiredhalf 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 outE 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]
Macsen wrote: 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. 3/36C. 11/36D. 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:6,22,64,45,33,5The 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 = 20/36Answer: D _________________
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Re: Two dice are tossed once. The probability of getting an even number at [#permalink] We van just list out the possibilites.2,12,22,32,42,52,63,54,14,24,34,44,54,65,36,16,26,36,46,56,6So 20/36. 18 cases occur from the first dice being even. The other two cases are 3,5 and 5,3 Posted from my mobile device
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Re: Two dice are tossed once. The probability of getting an even number at [#permalink]
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 isA. 1/36 B. 3/36C. 11/36D. 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. 3/36C. 11/36D. 20/36E. 23/361. 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.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 [#permalink]
Macsen wrote: 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. 3/36C. 11/36D. 20/36 E. 23/36 total possible outcomes for two dice: 36Prob of getting an even number on the first die 20/36
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Re: Two dice are tossed once. The probability of getting an even number at [#permalink]
Solution:For two events A and B,P(A or B)=P(A) + P(B) - P(A & B)P(A)=The probability of getting an even number at the first die = 3/6 = 1/2 {2,4,6 are the even numbers}P(B)=The probability of getting a total of 8 = 5/36 {(2,6)(3,5)(4,4),(5,3),(6,2) with a sample space of 6^2=36}The probability of getting an even number at the first die AND a total of 8 = 3/36 {(2,6),(4,4),(6,2)}=>The probability of getting an even number at the first die OR a total of 8 = P(A) + P(B) - P(A & B)= 1/2+5/36-3/36 =20/36 (option d) Devmitra Sen
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Two dice are tossed once. The probability of getting an even number at [#permalink]
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 isA. 1/36 B. 3/36C. 11/36D. 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. 3/36C. 11/36D. 20/36E. 23/361. 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.Answer: D. Hello Bunuel,In your explanation of the ''AND'' probability for the independent events, you mentioned that P (A and B) = P(A) X P(B), however, why this formula is not getting applied in the below question. If I try to apply this formula to find P (A intersection B) = 1/2 X 5/36= 5/72. However, this value obtained is different from the one that we received by manually calculating the number of favourable outcomes (i.e. 3/36)Can you please explain why this difference in the value?RegardsVighnesh
Two dice are tossed once. The probability of getting an even number at [#permalink] 07 Jan 2022, 05:55 |
What is the probability of getting an even sum on rolling 2 dice together both of which
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