Question

The manufacturer of a fertilizer guarantees that, with the aid of the fertilizer, 65% of planted seeds will germinate. Suppose the manufacturer is correct. If 6 seeds planted with the fertilizer are randomly selected, what is the probability that more than 4 of them germinate?

Carry your intermediate computations to at least four decimal places, and round your answer to at least two decimal places.

Answer 1

that means five or six right?
use the binomial for this one

Answer 2

I want to do it this way:
\[\frac{ 4 }{ 6} * \frac{ 65 }{ 100 }\]
but im sure that's too easy/wrong ??

Answer 3

\[P(x=5)=\binom{6}{5}.65^\times .35\\
P(x=6)=.65^6\]

Answer 4

I'm confused on the "x=5" thing, since 5 isn't part of the equation. (Sorry! Math is my worst subject.)

Answer 5

forget the \(x=5\) then

Answer 6

that just means the probability you get five successes

Answer 7

Just saw your "that means 5 or 6" comment. I get how you did the x=5 portion now.. I am still confused on the formulas for these.

Answer 8

lets compute the easy one first

Answer 9

what is the probability you get all six? the probability of getting any one success is \(.65\) and assuming the trials are independent, (which we do) then the probability of getting a success 6 times in a row means
got it on the first try AND on the second try AND on the third try AND on the fourth try AND on the fifth try AND on the sixth
i.e.
\[.65\times.65\times .65\times .65\times .65\]more succinctly written as \(.65^6\)

Answer 10

i capitalized AND

Answer 11

because that means multiply

Answer 12

I'm getting .0754 for that one

Answer 13

ok fine looks good
now what about the probability you get 5 out of 6?

Answer 14

that one does make sense to me. but for the 5 would we not just do \[.65^{5}\]
?

Answer 15

which i confused you by writing \(P(x=5)\)

Answer 16

oh no
you need a \(.65^5\) because that is
success AND success AND success AND success AND success

Answer 17

but what about the failure?

Answer 18

success AND success AND success AND success AND success AND failure

Answer 19

\[.65^5\times .35\] if they occur in that order
but they could occur in any of 6 different ways
SSSSSF
SSSSFS
SSSFSS
SSFSSS
SFSSSS
FSSSSS

Answer 20

so you have to take
\[.65^5\times .35\] and multiply it by \(6\) i.e.
\[6\times .65^5\times .35\]

Answer 21

I'm so sorry to ask this.. where do you get .35 from?

Answer 22

i will ask you
where do you think it came from?
hint: success AND success AND success AND success AND success AND \(\color{red}{\text{failure}}\)

Answer 23

well I tried doing 1/6 but i got .16 (.17 while rounding)

Boy i really got myself in over my head with this stats class!

Answer 24

there is no 1/6 in the problem
if the probability of success is .65 then the probability of failure is 1-.65=.35

Answer 25

your final answer is
\[.65^6+6\times .65^5\times .35\]

Answer 26

I'm getting 0.3191 (if rounding)

Answer 27

Thank you for explaining the .35 thing btw. It clicked immediately after you said it, i have a habit of over-complicating things in my head.