Question

Assume that n = 15, and p =7/10.
Find the probability of at least 2 successes and at least 3 failures.________________???

Answer 1

p tends to mean success

(p+q)^n i believe is the general formation

Answer 2

\[\binom{15}{2}.7^2.3^8\]
maybe? its been awhile

Answer 3

.3^3

Answer 4

its a little more than that since we got "at least" to deal with

Answer 5

would the root be higher

Answer 6

\[\binom{n}{r}p^r q^{n-r}\]

for "r"; then youd have to add in alot more stuff; maybe best to do a converse of it?

Answer 7

at most 1 and at most 2 and then flop it

Answer 8

http://stattrek.com/lesson2/binomial.aspx

that helps a little :) at least it lets me know i got the formula right in the end lol

Answer 9

Im still not getting the right answer....
these are killing me I do everything it says and it hardly works

Answer 10

you doing by calculator or by table?

Answer 11

calculator

Answer 12

15c0 p^0 q^15 is P(0)

15c1 p^1 q^14 is P(1)

15c2 p^2 q^13 is P(2)

i believe its best to work these first

Answer 13

then add them up?

Answer 14

P(0) = 1*1*.3^15 = 0
P(1) = 15*.7*.3^14 = 0
P(2) = 15C2*.7^2*.3^13 = 0

i keep getting rediculously small values for these

Answer 15

is this one problem...you want...

P(at least 2 successes and at least 3 failures)

?

Answer 16

yes it is one problem,

Answer 17

Zarkons gonna whip out a number thats right and easy to get im sure ;)

Answer 18

P(at least 2 successes and at least 3 failures)
=P(at least 2 successes and at most 12 successes)

\[=\sum_{k=2}^{12}{15\choose k}\left(\frac{7}{10}\right)^k\left(\frac{3}{10}\right)^{15-k}\]

Answer 19

i get \[\frac{174634353763317}{200000000000000}\]

Answer 20

\[\approx.873171768817\]

Answer 21

The complement would be quicker if your calculator isn't real nice ;)

My calculator is pretty nice so I did it my way