two different dice are tossed together find the probability (1) of getting a doublet and (11) of getting a sum of 10 of the numbers on the two dice?
Answers
Answered by
1115
Given that two different dice are tossed.
Total number of outcomes = 6^2
= 36.
(1) Let A be the event of getting a doublet.
n(A) = {1,1},{2,2},{3,3},{4,4},{5,5},{6,6}
= 6.
Required probability P(A) = 6/36
= 1/6.
(ii) Let B be the event of getting a sum of 10 of the numbers of two dice.
n(B) = {6,4},{4,6},{5,5}
= 3.
Required probability P(B) = n(B)/n(S)
= 3/36
= 1/12
Hope this helps!
Total number of outcomes = 6^2
= 36.
(1) Let A be the event of getting a doublet.
n(A) = {1,1},{2,2},{3,3},{4,4},{5,5},{6,6}
= 6.
Required probability P(A) = 6/36
= 1/6.
(ii) Let B be the event of getting a sum of 10 of the numbers of two dice.
n(B) = {6,4},{4,6},{5,5}
= 3.
Required probability P(B) = n(B)/n(S)
= 3/36
= 1/12
Hope this helps!
Answered by
251
Given:
Two dice
To find:
The probability of getting a double
The probability of getting a sum of 10
Solution:
Sample space = ( 1, 0 ), ( 1, 1 ), ( 1, 2 )... ( 6, 6 )
Hence,
Sample space = 36
To find the probability of getting a doublet,
Outcomes = ( 1, 1 ), ( 2, 2 ).. ( 6, 6 )
Possible outcomes = 6
P (getting a double) = 6 / 36
Hence, P (getting a double) = 1 / 6
To find the probability of getting a sum of 10,
Possible outcomes = ( 4, 6 ), ( 6, 4 ), ( 5, 5 )
P (getting a sum of 10 ) =3/36
Hence, P ( getting a sum of 10 ) = 1 / 12
Read more on Brainly.in - https://brainly.in/question/3098561
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