# studying; i still don't understand how to calculate this-binomial distribution

• October 24th 2007, 02:29 PM
beetz
studying; i still don't understand how to calculate this-binomial distribution
P(X< 3 n=15, successes=.4)
• October 24th 2007, 03:02 PM
Plato
Quote:

Originally Posted by beetz
i still don't understand how to calculate this-binomial distribution: P(X< 3 n=15, successes=.4)

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 $(0.6)^{15}$. That is because P(success) $=0.4$ while P(failure) $={1-(0.4)}=0.6$.

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

“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. $P(X = 2) = {{15} \choose 2}\left( {.04} \right)^2 \left( {0.6} \right)^{13}$

$P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)$
• October 24th 2007, 03:12 PM
beetz
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.
• October 24th 2007, 03:44 PM
Plato
That is correct. I get 0.027114000777216
• October 24th 2007, 04:08 PM
beetz
oh, ok. sweeeeeeet.

thanks for you help. i greatly appreciate it! :]
• October 24th 2007, 04:13 PM
beetz
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?