probablility (1 Viewer)

[asl2]

probability

A die is thrown three times, Find the probability of

i) three sixes
ii) no sixes
iii) two sixes
vi) at least two sixes

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is there any better way rather than drawing up a tree diagram... which will give u lots and lots of possibiilities and is also time consuming...

thx

 

[kini mini]

Just removed your typo above

The way to do it is to look at what the tree diagram represents and ask yourself how this might be done more efficiently.

Taking the example of the die, with each roll you add 6 branches to every pre-existing branch to describe the possibility of that sequence of events. A far better way to do it is to let the probability of a 6 be p(6) = 1/6 - this assumes that each event is equally likely to occur, that is all branches are the same. There are 6 faces of the die, hence 1/6.

So the probability of 3 sixes from 3 throws is (1/6)^3 x 1. The x 1 is the number of ways this can happen, in this case there's only 1. No sixes is then obvious.

Two sixes requires (1/6)^2 x (5/6) x 3 - there are 3 ways this can happen.

AT LEAST should always trigger an alarm in your brain because you may be able to do the problem more easily by subtracting events that do not satisfy the condition. But in this case all you have to do is add the probabilities of 2 sixes and 3 sixes, which you have already calculated .

FWIW I learned the above method before I was forced to construct trees, they always pissed me off too!

 

[Ultimate]

Re: probability

Originally posted by asl2
A die is thrown three times, Find the probability of

i) three sixes
ii) no sixes
iii) two sixes
vi) at least two sixes

-------------------------------

is there any better way rather than drawing up a tree diagram... which will give u lots and lots of possibiilities and is also time consuming...

thx

Just use the binomical probability theorem;
In answer to your q's,

i) P (3 six's, with die thrown 3 times) = 3C3 x (1/6)^3 = 1/216
ii) P (no sixes) = 3C0 x (1/6)^0 x (5/6)^ 3 = 125/216.
iii) P (2 six's) = 3C2 x (1/6)^2 x (5/6) = 3 x 1/36 x 5/6 = 5/72
iv) P(atleast 2 sixes) = P (2 six's) + P (3 sixes) = 5/72 + 1/216

hope that helps

 

[needtostudy]

use combination permutation rules.

 

[asl2]

i get the first 3 parts...


i three sixes... 1/6 * 1/6 * 1/6
ii no sixes... 5/6 * 5/6 * 5/6....
iii i used tree diagram for this part...


part vi... how do i do that... i didn't understand the solutions previously posted... and i don't know bionomial theory either !!!

thx

 

[kini mini]

I tried to avoid binomial theory, it's probably best learned systematically from your teacher instead of bits and pieces from us.

What I said about "at least two", was that you can sum the probability of three sixes and the probability of two sixes to satisfy the "at least" condition. Do you understand what I'm saying here?