Edexcel S2 - 27th June 2016 AM

5. Four coins are flipped together and the random variable H represents the number of headsobtained. Assuming that the coins are fair,
Given that the four coins are all biased such that the chance of each one showing a head is50% more than the chance of it showing a tail,
(d) find the probability of obtaining more heads than tails when the four coins are flippedtogether.


how would you guys do this? i'm so confuuzzled

Exactly, Idk? I thought it would be 3/4 but the mark scheme says 3/5

The probabilities are in a ratio of 1.5:1, or 3:2. Therefore 1/5 = 0.2 and probability of a head is 3 * 0.2 = 3/5.

yeah but why (((((((((((((((((((((((((((((((((((((((((
I saw that in the mark scheme but i don't understand that ratio stuff

It says 50% MORE

Meaning that the probability of a head = 1.5* probability of a tail

(2/5)*1.5 = 3/5

Is this exam in the morning


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How many past papers should have you done before the exam?

Since I don't understand it i might just cram it.

So if it was 75% more likely woudl the ratio be 1.75:1 ? and 50% less likely would it be... 0.50:1??/

Yeah.

Good luck everyone! Hopefully the questions are worded in a nice way :P

To be fair, I would recommend watching some GCSE videos.

Imagine if you have john, dave and mick, with a bag of rice.
Splitting 6 rice portions in equal parts.

2 goes to john, 2 goes to dave and 2 goes to mick

I.e. 2 : 2 : 2

Then John gets 2 out of the 6, so 2/6, and so on for dave and mick

Can somebody explain how to do part D?

Depends how intelligent you are/What UMS you need to get.

Generally you should have done from jan 2010 all the way to jun 2015 including all the januaries inbetween

Skim read it, but it seems like it's just the probability that the time she has to wait before being served is greater than or equal to 8.

She's already been queing for 5 minutes and the max additional time she can queue for is another 3 minutes. 5+3=8

Yes exactly