[MOVED] Probability that random students watch TV show

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chris2525

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Sep 4, 2006
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A school survey found that 7 out of 10 students watch the television program "Lost". If four random students are chosen, what is the probability that three of them watch "Lost"?

I think that to get this, you must do the following:

. . .(7/10)(7/10)(7/10)(3/10) = 343/10000

The reason is because of the probabilty that one person watches "Lost" is 7/10, and three people have to watch it. And the probability that someone doesn't watch it is 3/10.

I think that I might be right, but I am not sure. Can anyone help me please?
Thanks :D
 
The probability that three of fours students watch lost is

WL is watch lost NWL is not Watch lost.

List all the combinations and use the and(x) or(+) rule.

P(WL) AND P(WL) AND P(WL) AND P(NWL)

OR

P (WL) AND P(WL) AND P(NWL) AND P(WL)

OR

P(WL) AND P(NWL) AND P(WL) AND P(WL)

OR

P(NWL) AND P(WL) AND P(WL) AND P(WL)

=7/10 x 7/10 x 7/10 x 3/10
+ 7/10 x 7/10 x3/10 x 7/10 ....

which will give you 4(7/10 ^3 x3/10) = 0.4116 = 4116/10000

I may be wrong but that's how I see it.
 
Alright thanks.
My only question though, why is that being multiplied by four?
 
The OR between them means you have to add it 4 times so you just mulitply it by 4.
 
Wanted criminal is correct.

\(\displaystyle \L\\\begin{pmatrix}4\\3\end{pmatrix}p^{3}q=4(0.70)^{3}(0.30)=\frac{1029}{2500}\approx{0.4116}\)
 
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