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probability help!! (1 Viewer)

survivor

Member
Joined
Nov 10, 2002
Messages
110
hey if anyone can help me out with these questions it would be great!! thankyou in advance if you can

1.) an integer n, where 1<= n <=299, is selected at random. what is the probability that it is
a. divisible by 5?
b. not a multiple of 5?


2.) A and B play a game of chance. A tosses two dice and wins if he scores a 5 or 7. B draws two discs, without replacement from a bag containing 5 discs numbered 1,2,3,4,5 and wins if the sum of the numbers drawn exceeds 6.
who has the greater chance of winning and by how much?


thankyou again...............
 

wogboy

Terminator
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Sep 2, 2002
Messages
653
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Sydney
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Male
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2002
1) Firstly, how many integers are there in total? 299. And how many integers divisible by 5? Well there's: 5,10,15,...,295 which total up to 295/5 = 59.

So the probability of the random integer being divsible by 5 is 59/299

The probability of the integer not being divisible by 5 is 1 - 59/299 = 240/299

2) The probability of A winning is:

There are 6*6=36 possible ways of rolling two dice. How can person A win? If he rolls: (1,4) (2,3) (3,2) (4,1) (1,6) (2,5) (3,4) (4,3) (5,2) (6,1) --> 10/36 chance of winning
= 5/18 chance of winning.

The probability of B winning is:

There are 5*4=20 ways to draw these discs. He will win if he draws: (2,5) (3,4) (3,5) (4,3) (4,5) (5,2) (5,3) (5,4) --> 8/20 chance of winning
= 2/5 chance of winning.

You can see who has the greater chance of winning. :)
 

survivor

Member
Joined
Nov 10, 2002
Messages
110
thankyou
one more question
if u have for example...
results of an election were as follows- party x 60%
- party y 40%
and then 2 voters are selected at random(ie from either party x or y)
does the percentage change, like when you do a prob tree , for the second draw???
 

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