S2 - Binomial distrubution

A coin is biased so that a head is twice as likely to occur as is a tail. The coin is tossed repeated. Find the probability that:

a) The first tail will occur on the fifth loss.
b) In the first seven tosses, there will be exactly 2 tails.

So, probability of a head is 2/3 and probability of a tail is 1/3.

How do I do this binomially?

I don't think I can as I don't know the number of observations.

How else do I do it?

Reply 1

rec53

In the second part the coin tossed 7 times. Does that give you any clue as to how many observations we are doing?

The first part, if the first tail occurs on the fifth toss, what do we need to get on the first four tosses? Each event is independent, so what can we do with probabilities?

Reply 2

CalculusMan

OP

10

Ah of course, I was looking too far into it.

Another one:

Records kept in a hospital show that 3 out of every 10 casualties who come to the casualty department have to wait more than half an hour before receiving medical attention. Find to 3 dp, the probability of the first 8 casualties who come to the casualty department: (a) none, (b) more than two will have to wait more than half an hour before receiving medical attention. Find also the most probable number of the 8 casualties that will have to wait more than half an hour.

I have written: X ~ Bin(10, 0.3) .... Is this correct?

I keep confusing what 10 is and what 8 is...if that makes sense.

Reply 3

gyrase

3

CalculusMan

Records kept in a hospital show that 3 out of every 10 casualties who come to the casualty department have to wait more than half an hour before receiving medical attention. Find to 3 dp, the probability of the first 8 casualties who come to the casualty department: (a) none, (b) more than two will have to wait more than half an hour before receiving medical attention. Find also the most probable number of the 8 casualties that will have to wait more than half an hour.

I have written: X ~ Bin(10, 0.3) .... Is this correct?

I keep confusing what 10 is and what 8 is...if that makes sense.

X ~ Bin(8, 0.3)

a) P(X = 0)
b) P(X > 2) = 1 - (P(X = 1) + P(X = 2) + P(X = 0))

Reply 4

danniella

4

CalculusMan

Ah of course, I was looking too far into it.

Another one:

Records kept in a hospital show that 3 out of every 10 casualties who come to the casualty department have to wait more than half an hour before receiving medical attention. Find to 3 dp, the probability of the first 8 casualties who come to the casualty department: (a) none, (b) more than two will have to wait more than half an hour before receiving medical attention. Find also the most probable number of the 8 casualties that will have to wait more than half an hour.

I have written: X ~ Bin(10, 0.3) .... Is this correct?

Sorry... no. X ~Bin(8, 0.3)

I keep confusing what 10 is and what 8 is...if that makes sense.

If 3 out of every 10 casualties have to wait more than half an hour, then that means that the probability, p, of someone having to wait is 0.3.

In your question, there are 8 people/casualties involved. So n = 8.

So X = no of casualties who have to wait over half an hour
x~Bin(8, 0.3)

Then a) is asking you to find P(X=0)
P(X=0) = ⁸C₀ 0.3^0 x 0.7^8

b) is asking you to find P(X>2)
If you need more than two people to wait over half an hour, then the easiest way to work out that probability is to find P(2 or less people waited over half an hour) and subtract that from 1.
So you work out P(X is less than or equal to 2) and then find
1-P(X less than or equal to 2)

P(x less than or equal to 2) = P(X=0) + P(X=1) + P(X=2)
= ⁸C₀ 0.3^0 x 0.7^8
+ ⁸C₁ 0.3^1
+ ⁸C₂ 0.3^2 x 0.7^6

and then subtract whatever that answer is from 1.

love danniella

Reply 5

danniella

4

CalculusMan

Find also the most probable number of the 8 casualties that will have to wait more than half an hour.

Oops, forgot this bit.
The most probable number of the 8 casualties that will have to wait over half an hour is the mean of the distribution, which is np (n= no of caaulaties, 8, and p = 0.3, the probability that someone has to wait more than half an hour).

So if X~Bin (8, 0.3) then E(X) = mean = 8x0.3 = 2.4 which would mean that 2 people would have to wait more than half an hour.

love danniella

S2 - Binomial distrubution

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