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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
A. 1/36
B. 3/36
C. 11/36
D. 20/36
E. 23/36
24 Jan 2012, 02:47
Kudos
Bookmarks
Macsen
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.
Answer: D.
23 Jan 2012, 21:58
Kudos
Bookmarks
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
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
