The Student Room Group

S1 probability question

Reply 1


Have you heard of conditional probability?
0you have to use the conditional probability expansion:
P(A/B)=P(A and B) over p(B)
Reply 3
P(car owner given bike owner) = P( car and bike owner) / P(bike owner)
Reply 4
Original post by Zacken
Have you heard of conditional probability?


yeah my working for part c is :
P(car owner/bike owner)
= 0.78 / (0.85+0.3)
= 0.678
I know it is wrong and i dont know how i should do it
Reply 5
Original post by alesha98
yeah my working for part c is :
P(car owner/bike owner)
= 0.78 / (0.85+0.3)
= 0.678
I know it is wrong and i dont know how i should do it


Look up the formula again. P(car owner AND bike owner) = 78 * 0.3 should be in the numerator.

The denominator is clearly wrong, you can't have a probability of greater than 1.

The probability of being being a bike owner is either being a car owner and a bike owner (0.85 * 0.3) or (+) not a car owner and a bike owner (0.22 * 0.85).

Have you drawn a tree diagram...?
Reply 6
Original post by Zacken
Look up the formula again. P(car owner AND bike owner) = 78 * 0.3 should be in the numerator.

The denominator is clearly wrong, you can't have a probability of greater than 1.

The probability of being being a bike owner is either being a car owner and a bike owner (0.85 * 0.3) or (+) not a car owner and a bike owner (0.22 * 0.85).

Have you drawn a tree diagram...?


yes i have drawn a tree diagram and sorry for the non knowledgeable reply, but i am really bad in stats...
My tree diagram is right and after i have read your comment, i think i realised i should have done:
0.78 x 0.3 / 0.78 x 0.3 + 0.22 x 0.85 ?
Reply 7
Original post by alesha98
yes i have drawn a tree diagram and sorry for the non knowledgeable reply, but i am really bad in stats...
My tree diagram is right and after i have read your comment, i think i realised i should have done:
0.78 x 0.3 / 0.78 x 0.3 + 0.22 x 0.85 ?


Yes, that's correct.
Reply 8
Original post by Zacken
Yes, that's correct.


would u mind tell me how to do part d as well?
Reply 9
Original post by alesha98
would u mind tell me how to do part d as well?


What's the probability of someone being a bike owner? Look at your tree diagram and multiply along the branches and add up the relevant probabilities.

So, what is the probability that the person is not a bike owner? (hint: 1- probability he is a bike owner).

Then the probability that only one person is a bike owner amongst 2 people is:

P(Bike owner)P(not bike owner) + P(not bike owner)P(bike owner).
Reply 10
Original post by Zacken
What's the probability of someone being a bike owner? Look at your tree diagram and multiply along the branches and add up the relevant probabilities.

So, what is the probability that the person is not a bike owner? (hint: 1- probability he is a bike owner).

Then the probability that only one person is a bike owner amongst 2 people is:

P(Bike owner)P(not bike owner) + P(not bike owner)P(bike owner).


the probability of a bike owner is 0.78 x 0.3 + 0.22 x 0.85 = 0.421
The probability of not a bike owner is 0.78 x 0.7 + 0.22 x 0.15 = 0.579
So The probability of one person out of two is a bike owner would be:
2(0.421 x 0.579)?
the highlighted bit i still dont understand
Reply 11
Original post by alesha98
the probability of a bike owner is 0.78 x 0.3 + 0.22 x 0.85 = 0.421
The probability of not a bike owner is 0.78 x 0.7 + 0.22 x 0.15 = 0.579
So The probability of one person out of two is a bike owner would be:
2(0.421 x 0.579)?
the highlighted bit i still dont understand


Did you read my post? What does the last line of my post say?
Reply 12
Original post by Zacken
Did you read my post? What does the last line of my post say?


P(Bike owner)P(not bike owner) + P(not bike owner)P(bike owner).
I dont get it ...
Reply 13
Original post by alesha98
P(Bike owner)P(not bike owner) + P(not bike owner)P(bike owner).
I dont get it ...


Well, you have two people. Let's call them Alice and Bob. You want only one of them to own a bike. If one owns a bike, the other shouldn't own a bike.

So P(Alice owns a bike) * P(Bob does not own a bike) is one possibility for only one of the the two people to have a bike.

P(Bob owns a bike) * P(Alice does not own a bike) is the other possibility for only of of the two people to have a bike.

So you add the above two.
Reply 14
Original post by Zacken
Well, you have two people. Let's call them Alice and Bob. You want only one of them to own a bike. If one owns a bike, the other shouldn't own a bike.

So P(Alice owns a bike) * P(Bob does not own a bike) is one possibility for only one of the the two people to have a bike.

P(Bob owns a bike) * P(Alice does not own a bike) is the other possibility for only of of the two people to have a bike.

So you add the above two.

stats really does confused me... Thankyou and i think i got this question now
Reply 15
Original post by alesha98
stats really does confused me... Thankyou and i think i got this question now


Awesome.

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