probability help. (1 Viewer)

[kolgf]

I am having trouble with these questions and i need to noe how to work them out:

1. Kate organises a household roster for her family so that every night one person washes up while another wipes up. There are 6 people in her family: Mum, Dad and 4 children.

( a ) how many pairings r possible? (answer = 30)
( b ) How many pairings are possible if dad woun't wipe? (answer=25)
( c ) How many pairings are possible if dad woun't wipe and mum wount wash up? (answer=21)

thanks for any help.

 

[Riviet]

a) Mum can be paired with each of the other 5 in the family, in which for each pair, one could do washing, the other wipe up and they could swap roles so we multiply the number of pairs by two: 2 x 5 = 10pairs. Notice how we have counted all the possible combinations that mum could pair up with the other 5, so we don't need to worry about her anymore.

Similarly for the 5 other members, Dad can pair up with the 4 kids, with each pair having two different combinations, just like before with one washing, one wiping, and their roles swapped, so there's 2 x 4 = 8 ways. Notice how we have counted all the ways that Dad can pair up with the kids, and remembering that we already counted Mum and Dad as a pair before.

Similarly for the reminaing 4 kids, one of them can pair up with the other 3 in 3 x 2 =6 ways. Same thing for the remaining 3, 2 kids which have 2x2=4 and 2x1=2 ways of pairing up respectively.

So total number of ways = 10+8+6+4+2 = 30 ways.

b) Since dad won't wipe, we find the number of pairings in which Dad DOES wipe and subtract this from the total, which we found in part (a) to be 30.

Dad can be paired up in 5 ways with each of the other 5 family members, so we subtract 5 from 30 to get 25 ways.

c) Using part (b), there were 5 pairs in which Dad wouldn't wipe. Similarly there are 5 pairs in which mum won't wash up, BUT we have already counted the pair in which mum wipes and dad washes up, so we only count the 4 pairs that mum can make with the 4 kids.

So we subtract these from the total number to find the number of pairings possible: 30-4-5=21.

 

[kolgf]

Makes sense, thanks alot Riviet.

 

[PC]

Another way to figure out these questions is to use a tree diagram in each case. It's not as succinct as Riviet's answers, but it will get you the answers.

 

[Dr_Doom]

This took us ages to work out in class.

 

[kolgf]

i got another probility problem how do u figure this out ( i am not good in this topics ^_^)

1. Tegan buys 5 tickets in a raffle in which 80 tickets are sold. There are 3 prizes. Using a tree diagram find the probability that tegan :

a) wins all 3 prizes
b) does nto win a prize
c) wins atleast 1 prize
d) wins exactly one prize

i dnt get it

 

[pLuvia]

a)P(wins all 3 prizes)=(5/80*4/79*3/78)=1/8216
b)P(does not win a prize)=(75/80*74/79*73/78)=0.82
c)P(wins at least 1 prize)=P(1-[not winning a prize])=1-0.82=0.18
d)P(exactly one prize)=(5/80*75/79*74/78+75/80*5/79*74/78+75/80*74/79*5/78)=0.17

 

[kolgf]

thanks, is there anyway that i can learn probility, coz its the only subject i dnt get no matter how much i study it.

 

[pLuvia]

It's the matter of just doing heaps of questions and finding shortcuts eg Q(c). Helpful if you draw a probability tree even though it is sometimes annoying