#### XcarvengerX

##### Chocobo

So the paper has 9 marks on probability, but they are quite easy. These are the correct solutions.

3.(i)

(ii)

6.

(i) Binomial distribution using

Answer:

(ii) 1 - (

= 1 - 4q

(iii) Substituted p=1-q to

= 1 -

= 1 - q

(iv) P

1 - q

= 1/3 < q < 1

Thanks to pesila for typing the workout, but thanks to me for all the subscripts and supscripts.

3.(i)

^{5}P_{3}= 60(ii)

^{5}P_{2}+^{5}P_{3}+^{5}P_{4}+^{5}P_{5}= 3206.

(i) Binomial distribution using

^{4}C_{3}and^{4}C_{4}Answer:

^{4}C_{4}q^{4}+^{4}C_{3}q^{3}p(ii) 1 - (

^{4}C_{4}q^{4}+^{4}C_{3}q^{3}p)= 1 - 4q

^{3}+ 4q^{4}(iii) Substituted p=1-q to

^{2}C_{1}pq +^{2}C_{2}p^{2}= 1 -

^{2}C_{2}q^{2}= 1 - q

^{2}(iv) P

_{(team of 2 scores points)}> P_{(team of 4 scores points)}1 - q

^{2}> 1 - (4q^{3}- 3q^{4})= 1/3 < q < 1

Thanks to pesila for typing the workout, but thanks to me for all the subscripts and supscripts.

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