Question

Compute the following binomial probabilities directly from the formula for b(x; n, p):

a. b(3; 8, .35)
b. b(5; 8, .6)
c. P(3<=X<=5) when n=7 and p = .6
d. P(1 <= X) when n=9 and p= .1

Answer 1

8c3 p^3 q^5

Answer 2

p = .35
q = 1-p = .65

Answer 3

for b. its the same process

8c5 p^5 q^3

Answer 4

c is to find the value at 5 and subtract the value from 3

Answer 5

or rather; 3+4+5

Answer 6

i was confusing cumulative with point :)

Answer 7

does any of this ring a bell?

Answer 8

Thanks for replying, amistre. I don't understand how to use the formula :(

Answer 9

the formula deals with factorial/combonatoric notation

Answer 10

we find the binomial coefficient which tells us how many times the particular set up can occur.

8! 8.7.6.5.4.3.2 8.7.6
----- = ------------- = ---- = 8.7 = 56
3!5! 3.2 5.4.3.2 3.2

the even occurs 56 times; now we have to just determine the probability of a single event

Answer 11

a single event that has the set up of 3 successes (p) and the rest failures (q) is the product of p.p.p and q.q.q.q.q

Answer 12

since p = .35; q is the complement of it; .65

.35*.35*.35* .65*.65*.65*.65*.65*

(.35)^3 * (.65)^5

Answer 13

all together we get:

56 * (.35)^3 * (.65)^5

Answer 14

that is the basis for the formula; now if any of it doesnt make sense, just say so

Answer 15

and just to chk:

56 * (.35)^3 * (.65)^5 = .27858...

and the table from the wolf gives us: 3 = .2786

Answer 16

ok, I get it now. The examples in my textbook were throwing me off, according to which I was going to do (8 choose 3) * (.35)^3 * (.35)^3

Answer 17

that would have been close :)

that last bit tho is "whats left over" from the success. in other words:

\[\Large ^8C_3\ * p^3\ * (1-p)^{8-3}\]

Answer 18

b(5; 8, .6); p=.6, so q=.4

\[\Large ^8C_5\ * p^5\ * q^{8-5}\]

Answer 19

Gives you same answer as part a?

Answer 20

P(3<=X<=5) when n=7 and p = .6; q=.4

\[\Large ^7C_3\ * p^3\ * q^{7-3}\]
\[\Large +^7C_4\ * p^4\ * q^{7-4}\]
\[\Large +^7C_5\ * p^5\ * q^{7-5}\]

Answer 21

... urghh. let me try that again but with the actual problem instead

Answer 22

56 * .6^5 * .4^3 = .2787

so yeah :)

Answer 23

the process is the same; it just so happens that the answers are a close fit by coincidence

Answer 24

awesome! and I understand part c from what you did above. Thanks so much for explaining!

Answer 25

P(1 <= X) when n=9 and p= .1; q=.9

\[\Large ^9C_0\ * p^0\ * q^{9-0}\]\[\Large +^9C_1\ * p^1\ * q^{9-1}\]

youre welcome :)

Answer 26

do you see my error in the last part here?

Answer 27

i read X as less than or equal to 1; and summed up the probabilities for the complement instead; which is less work and fixable

Answer 28

if we are looking for a value of say: A and we find its complement: A'; then since A + A' = 1 we can find A

A = 1 - A'

Answer 29

let me know if that makes sense :)

Answer 30

so would my answer for part d be approximately .7748?

Answer 31

ok, it makes sense. Thanks again amistre!!