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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
Correct answer is option 'D'. Can you explain this answer?
Verified Answer
. Two dice are tossed once. The probability of getting an even number ...
A = getting even no on 1st dice
B = getting sum 8
B = getting sum 8
Most Upvoted Answer
. Two dice are tossed once. The probability of getting an even number ...
Solution:
Possible outcomes when two dice are tossed once = 6 × 6 = 36
Let E1 be the event of getting an even number on the first die
Let E2 be the event of getting a total of 8
P(E1) = probability of getting an even number on the first die
= 3/6 (as three even numbers are possible on a die)
= 1/2
P(E2) = probability of getting a total of 8
= number of possible outcomes to get a total of 8 / total possible outcomes
= 5/36
(we get a total of 8 when we get (2,6), (3,5), (4,4), (5,3), or (6,2))
P(E1 ∪ E2) = probability of getting an even number on the first die or a total of 8
= P(E1) + P(E2) - P(E1 ∩ E2)
(where P(E1 ∩ E2) is the probability of getting an even number on the first die and a total of 8, which is not possible)
= 1/2 + 5/36
= 20/36
Therefore, the probability of getting an even number on the first die or a total of 8 is 20/36 or option D.
Possible outcomes when two dice are tossed once = 6 × 6 = 36
Let E1 be the event of getting an even number on the first die
Let E2 be the event of getting a total of 8
P(E1) = probability of getting an even number on the first die
= 3/6 (as three even numbers are possible on a die)
= 1/2
P(E2) = probability of getting a total of 8
= number of possible outcomes to get a total of 8 / total possible outcomes
= 5/36
(we get a total of 8 when we get (2,6), (3,5), (4,4), (5,3), or (6,2))
P(E1 ∪ E2) = probability of getting an even number on the first die or a total of 8
= P(E1) + P(E2) - P(E1 ∩ E2)
(where P(E1 ∩ E2) is the probability of getting an even number on the first die and a total of 8, which is not possible)
= 1/2 + 5/36
= 20/36
Therefore, the probability of getting an even number on the first die or a total of 8 is 20/36 or option D.
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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 isa)1/36b)3/36c)11/36d)20/36Correct answer is option 'D'. Can you explain this answer?
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