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junior college 2 | H2 Maths
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Could anyone help to explain why P(A) is 30/36? I thought it would simply be 5/6 given that there are 5 outcomes that are not 6.
Even though 30/36 still simplifies to 5/6 eventually but I would like to understand the process.
And how to obtain P(A n B) = 25/36 prior to using 5/6 x 5/6.
Not exactly JC math but I thought still elementary enough for a JC tutor.
Thank you!
Date Posted: 5 years ago
Views: 465
So for the first 2 parts I think they were considering what die 2 will show as well!
So... For each number that die 1 shows, die 2 can show 6 possibilities! So in the end the number of combination is 36 and out of those only 6 of them are when die 1 is 6 and die 2 can be (1,2,3,4,5,6)
Hope that clears your doubt!
For the second one I think u can use this formula ~
P(A ∩ B) = P(A) P(B|A)
So P(A) =30/36=5/6
P(B|A) is the probability of B when A has occurred so... That will just be 5/6! :)
So... For each number that die 1 shows, die 2 can show 6 possibilities! So in the end the number of combination is 36 and out of those only 6 of them are when die 1 is 6 and die 2 can be (1,2,3,4,5,6)
Hope that clears your doubt!
For the second one I think u can use this formula ~
P(A ∩ B) = P(A) P(B|A)
So P(A) =30/36=5/6
P(B|A) is the probability of B when A has occurred so... That will just be 5/6! :)
See 2 Answers
Hahahhaha
done
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Hope it helps
Date Posted:
5 years ago
This is helpful & well explained! Thank you
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