2. Two dice are thrown together examples
1) Two dice are thrown together. What is the probability that numbers on the two faces sum/add is Even .
Solution:
Total number of outcomes possible when a die is thrown = 6,
Hence, Total number of outcomes possible when two dice are thrown is
`n(S)=6 xx 6=36`
Let `E` = event of getting a total is even.
`:.E = {(1,1), (1,3), (1,5), (2,2), (2,4), (2,6), (3,1), (3,3), (3,5), (4,2), (4,4), (4,6), (5,1), (5,3), (5,5), (6,2), (6,4), (6,6)}`
`:.n(E) = 18`
`:.P(E)=(n(E))/(n(S))=18/36=1/2`
2) Two dice are thrown together. What is the probability that numbers on the two faces product/multiple is = 4 .
Solution:
Total number of outcomes possible when a die is thrown = 6,
Hence, Total number of outcomes possible when two dice are thrown is
`n(S)=6 xx 6=36`
Let `E` = event of getting a product `=` 4.
`:.E = {(1,4), (2,2), (4,1)}`
`:.n(E) = 3`
`:.P(E)=(n(E))/(n(S))=3/36=1/12`
3) Two dice are thrown together. What is the probability that numbers on the two faces sum/add is >= 5 and sum/add is Prime
Solution:
Total number of outcomes possible when a die is thrown = 6,
Hence, Total number of outcomes possible when two dice are thrown is
`n(S)=6 xx 6=36`
Let `E` = event of getting a total `>=` 5 and total is prime.
`:.E = {(1,4), (1,6), (2,3), (2,5), (3,2), (3,4), (4,1), (4,3), (5,2), (5,6), (6,1), (6,5)}`
`:.n(E) = 12`
`:.P(E)=(n(E))/(n(S))=12/36=1/3`
4) Two dice are thrown together. What is the probability that numbers on the two faces sum/add is = 10 or sum/add is = 11
Solution:
Total number of outcomes possible when a die is thrown = 6,
Hence, Total number of outcomes possible when two dice are thrown is
`n(S)=6 xx 6=36`
Let `E` = event of getting a total `=` 10 or total `=` 11.
`:.E = {(4,6), (5,5), (5,6), (6,4), (6,5)}`
`:.n(E) = 5`
`:.P(E)=(n(E))/(n(S))=5/36`
5) Two dice are thrown together. What is the probability that numbers on the two faces doublet
Solution:
Total number of outcomes possible when a die is thrown = 6,
Hence, Total number of outcomes possible when two dice are thrown is
`n(S)=6 xx 6=36`
Let `E` = event of getting a doublet.
`:.E = {(1,1), (2,2), (3,3), (4,4), (5,5), (6,6)}`
`:.n(E) = 6`
`:.P(E)=(n(E))/(n(S))=6/36=1/6`
6) Two dice are thrown together. What is the probability that numbers on the two faces product/multiple is Divide by 3 and product/multiple is Divide by 5
Solution:
Total number of outcomes possible when a die is thrown = 6,
Hence, Total number of outcomes possible when two dice are thrown is
`n(S)=6 xx 6=36`
Let `E` = event of getting a product divide by 3 and product divide by 5.
`:.E = {(3,5), (5,3), (5,6), (6,5)}`
`:.n(E) = 4`
`:.P(E)=(n(E))/(n(S))=4/36=1/9`
7) Two dice are thrown together. What is the probability that numbers on the two faces sum/add is >= 5 and sum/add is <= 6
Solution:
Total number of outcomes possible when a die is thrown = 6,
Hence, Total number of outcomes possible when two dice are thrown is
`n(S)=6 xx 6=36`
Let `E` = event of getting a total `>=` 5 and total `<=` 6.
`:.E = {(1,4), (1,5), (2,3), (2,4), (3,2), (3,3), (4,1), (4,2), (5,1)}`
`:.n(E) = 9`
`:.P(E)=(n(E))/(n(S))=9/36=1/4`
This material is intended as a summary. Use your textbook for detail explanation.
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