Probability of winning four-digit state lottery

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sunny1324

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in a state lottery four digits are drawn at random one at a time with replacement from 0 to 9. Suppose that you win if any permutations of your selected integers is drawn. Give the probability of winning if you select

a. 6,7,8,9
b. 6,7,8,8
c. 7,7,8,8
d. 7,8,8,8,

help please. i don't even know where to get started.
 
Re: Probability of winning

sunny1324 said:
in a state lottery four digits are drawn at random one at a time with replacement from 0 to 9. Suppose that you win if any permutations of your selected integers is drawn. Give the probability of winning if you select

a. 6,7,8,9
b. 6,7,8,8
c. 7,7,8,8
d. 7,8,8,8,

help please. i don't even know where to get started.
Click to expand...

Start with calculating - how many four digit numbers can be drawn.
 
Re: Probability of winning

the total would be 10(10)(10)(10) = 10000

a. 6,7,8,9
4! = 24
24/10000 = 0.0024

that's the right answer, but i think it might be a coincidence. I still don't understand it.
 
Re: Probability of winning

sunny1324 said:
the total would be 10(10)(10)(10) = 10000
a. 6,7,8,9
4! = 24
24/10000 = 0.0024
that's the right answer, but i think it might be a coincidence. I still don't understand it.
Click to expand...
Why? What is to understand?
Do part b: 6788
\(\displaystyle {4 \choose 2} = 6\) is the number of ways to place the 8's, four choose two.
Double the to account for 6 & 7.
Is \(\displaystyle \frac {12} {10000}\) the correct answer?
 
Re: Probability of winning

that is the right answer, thanks


i get it now. :)
for c, it would be
4!/ (2!2!) = 6
6/10000
 
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