# Emergency probability help needed s1 exam tomorrow!!!

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#1
(ii) Show that the events ‘the score on the first die is even’ and ‘the product of the scores on the two
dice is less than 10’ are not independent.
ON THIS PAPER
http://mei.org.uk/files/papers/2011_Jan_s1.pdf
I just dont get it.
0
4 years ago
#2
Work out both probabilities, probabilities of both occurring, then times the two probabilities together and prove it doesn't equal the probability of both occurring.
0
4 years ago
#3
Revision notes:

For dependent events (one event that depend on the other) : P(AnB) = P(A) x P(B|A) (probability of B given A has occurred).

For independent events (events that occur without influence): P(AnB)=P(A) x P(B)
0
4 years ago
#4
If A and B are independent, P(B|A) = P(B|A') = P(B)

If they are dependent they don't equal as it depends on the outcome of A.

An example of independent would be the probability of me eating cereal or toast for breakfast and the probability of me passing the exam. They have no dependency on each other at all.

A dependent event would be the probability I forgot my calculator and the probability I passed the exam. The calculator forgetting would stress me out, id have to find one and I'd just panick too much, lowering my chances, there P(B) (me passing) would change in occurance to P(A) (me forgetting my calculator.)
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#5
(Original post by ExoIceCream99)
Work out both probabilities, probabilities of both occurring, then times the two probabilities together and prove it doesn't equal the probability of both occurring.
thank you but i dont know how to get the probability of both occurring p(AnB)
can you please explain it to me would be much appreciated.
I know how to get the first P(A)*P(B)
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4 years ago
#6
thank you but i dont know how to get the probability of both occurring p(AnB)
can you please explain it to me would be much appreciated.
I know how to get the first P(A)*P(B)
Look at my second comment
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#7
(Original post by ExoIceCream99)
Look at my second comment
but how do i work out BIA don't i need p(AnB)
sorry man im so bad at probability. Could you please just give a solution i have strong feeling this is gnna come up tmrw
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4 years ago
#8
but how do i work out BIA don't i need p(AnB)
sorry man im so bad at probability. Could you please just give a solution i have strong feeling this is gnna come up tmrw
P(B|A) = probability of B and A occurring / probability of A
0
#9
(Original post by ExoIceCream99)
P(B|A) = probability of B and A occurring / probability of A
is the probability of b and a occurring the P(B)*P(A)
0
4 years ago
#10
So if the probability of me having cereal is 0.8 P(A) and P(B) me passing is 0.6

P(B|A) = 0.8 x 0.6 / 0.8
P(B|A') = 0.2 x 0.6 / 0.2

If they equal the same (which in this case they both equal 0.6) then they are independent events.
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#11
(Original post by ExoIceCream99)
So if the probability of me having cereal is 0.8 P(A) and P(B) me passing is 0.6

P(B|A) = 0.8 x 0.6 / 0.8
P(B|A' = 0.2 x 0.6 / 0.2

If they equal the same (which in this case they both equal 0.6) then they are independent events.
Alright thank you so much good luck
for tomorrow.
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4 years ago
#12
Alright thank you so much good luck
for tomorrow.
You too!!!

You'll be fine
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#13
(Original post by ExoIceCream99)
You too!!!

You'll be fine
Thanks I hope so!
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