probability question (1 Viewer)

[beanari]

ok i suck at probability. please explain how this works..

Ryan buys three tickets in a guessing competition which has two prizes. Altogethr 100 tickets are sold.

calculate the probability that he will win
i. both first and second prize
ii. exactly one prize
iii. at least one prize
iv. no prize


help. i totally dont get this

 

[forevaunited]

ok i suck at probability. please explain how this works..

Ryan buys three tickets in a guessing competition which has two prizes. Altogethr 100 tickets are sold.

calculate the probability that he will win
i. both first and second prize
ii. exactly one prize
iii. at least one prize
iv. no prize


help. i totally dont get this

The easiest way is to draw a tree diagram. Remember the facts. THere are 2 prizes. THerefore on the first pick of a prize he has a 3 in 100 chance of winning. The chance of losing on the first pick is therefore 97/100.

Ok so if he wins the first prize it means that a ticket has been taken away. Therefore the chance of winning on the second go is 2 / 99 with losing being 97/99.

Ok so the questions

i.) Well first pick of the prizewinners says that he has a 3 in 100 chance. Then if he wins that - the chance of winning the second prize also becomes 2/99.

Therefore 3/100 * 2/99 = 1 / 1650

ii.) There are two prizes, so two chances of winning, so the possibe combinations for exactly 1 prize win is WL, LW.

That means (3/100 * 97/99) + (97/100 * 3/99) = 97 / 1650

iii.)At least one prize means that we can have: WW, WL, LW. Alternatively we can subtract the possibility of LL from 1.

P(At least one win) = 1- P(LL)

1 - (97/100 * 96/99) = 49 / 825

iv.) No prize is LL so,

97/100 * 96/99 = 776 / 825

 

[imaginarylife]

i) he has a 3 in 100 chance to win the first prize as he bought 3 tickets out of the 100 sold. To win the second prize as well he has a 3/100 x 2/99 chance. 3/100 because this is the chance he has to win first and 2/99 as he has 2 tickets left and overall there are 99 tickets left (as first prize has already been drawn). You times them together as both probabilities must be true in this question.

ii)this has two options for him winning only one prize, first, that he wins 1st prize, second, that he just wins 2nd prize. his chance of just winning first prize is 3/100 x 97/99 (his chance of winning 1st prize multiplied by his chance of not winning 2nd prize) for second prize it is 97/100 x 3/99 (his chance of not winning first x his chance of not winning 2nd) then you add the two together to get the overall probability. ie (3/100 x 97/99) + (97/100 x 3/99)

iii) for this you need to add together all the options he has of winning.
first of winning just one prize, and also of winning both prizes
ie. (3/100 x 97/99) + (97/100 x 3/99) + (3/100 x 2/99)

iv) for no prize then there is a 97/100 chance of not winning the first multiplied by 96/99 chance of not winning the second. ie. 97/100 x 96/99

you can use tree graphs to solve these questions

[please correct me if I've made a mistake] (ahh it seems that someone else posted while i was still writing mine, but it seems we have the same answers)

 

[Hallatia]

I find general maths really easy, so happy I moved down