P(X< 3 n=15, successes=.4)

- Oct 24th 2007, 02:29 PMbeetzstudying; i still don't understand how to calculate this-binomial distribution
P(X< 3 n=15, successes=.4)

- Oct 24th 2007, 03:02 PMPlato
Think of a string of letters 15 places long.

“FFFFFFFFFFFFFFF” that would be a string of 15 failures and can happen in only with probability $\displaystyle (0.6)^{15}$. That is because P(success) $\displaystyle =0.4$ while P(failure) $\displaystyle ={1-(0.4)}=0.6$.

“SFFFFFFFFFFFFFF” that would be a string of 14 failures and one success. It can happen in 15 ways (S can be in 15 places). $\displaystyle P(X = 1) = 15\left( {.04} \right)\left( {0.6} \right)^{14} $.

“SSFFFFFFFFFFFFF” that would be a string of 13 failures and two successes. It can happen in combination of 15 taken 2 at a time ways. $\displaystyle P(X = 2) = {{15} \choose 2}\left( {.04} \right)^2 \left( {0.6} \right)^{13} $

$\displaystyle P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)$ - Oct 24th 2007, 03:12 PMbeetz
hmmm...

so, 0 + .0047 + 105 ( 15*14/2)* (.4)^2(.6)^13

--- 0 + .0047 + .0219= .0266

doesn't seem right....did i calculate wrong?

do you know how to do this on a ti using the binom. functions?

thanks. - Oct 24th 2007, 03:44 PMPlato
That is correct. I get 0.027114000777216

- Oct 24th 2007, 04:08 PMbeetz
oh, ok. sweeeeeeet.

thanks for you help. i greatly appreciate it! :] - Oct 24th 2007, 04:13 PMbeetz
now, if the problem were, say, less than 10, it would be the same concept, but for x=2, 15/2 (.4)^2 (.6)^13 and x=3 15/3 (.4)^3(.6)^11 etc. all the way to x=9?