Some people believe in the force (yes, like in Star Wars). To test if young Anakin Skywalker has the force with him, he is told that some cards that Yoda can see but Anakin cannot contain a picture of a planet, a lightsaber, a laser rifle, or a spaceship. As Yoda looks at 20 such cards in turn, Anakin tries to guess what is on the card Yoda is looking at. Of course, Anakin has a 25% chance of simply guessing correctly.

a) Verify that the count of correct guesses in 20 cards follows a binomial distribution and write the notation.
b) What is the mean number of correct guesses?

Question

Some people believe in the force (yes, like in Star Wars). To test if young Anakin Skywalker has the force with him, he is told that some cards that Yoda can see but Anakin cannot contain a picture of a planet, a lightsaber, a laser rifle, or a spaceship. As Yoda looks at 20 such cards in turn, Anakin tries to guess what is on the card Yoda is looking at. Of course, Anakin has a 25% chance of simply guessing correctly.

a) Verify that the count of correct guesses in 20 cards follows a binomial distribution and write the notation.
b) What is the mean number of correct guesses?

anonymous · Answer

c) What is the probability that Anakin guesses all 20 cards correctly?
d) Suppose Anakin guesses correctly on 10 of the cards. What is the probability of him doing this well or better by chance? Do you think he has the force with him?

anonymous · Answer

@amistre64

jamesj · Answer

Let p for correct be equal to the probability of guessing one card correctly. Then p = 1/4.  The probability of getting one card wrong is 1 - p = 1 - 1/4 = 3/4.

So far so good?

jamesj · Answer

talk to me, or I'll go away

anonymous · Answer

@JamesJ I am not always on this website.... SOrry. but im here now, are you

anonymous · Answer

What is the mean number of correct guesses?

have a one in four chance of getting it right, so $20\times \frac{1}{4}$

anonymous · Answer

5

anonymous · Answer

c) What is the probability that Anakin guesses all 20 cards correctly?

to get 20 in a row he guesses correctly each time
in one guess it is $\frac{1}{4}$ to get two in a row it would be $\frac{1}{4}\times \frac{1}{4}=\left(\frac{1}{4}\right)^2$

anonymous · Answer

to get three in a row: $\left(\frac{1}{4}\right)^3$ and so on

anonymous · Answer

Did u find a or b?

anonymous · Answer

d) Suppose Anakin guesses correctly on 10 of the cards. What is the probability of him doing this well or better by chance? Do you think he has the force with him?

since you have verified this is in fact binomial, use 
$$P(x=k)=\binom{n}{k}p^k(1-p)^{n-k}$$ with 
$$n=20,k=10, p=\frac{1}{4}$$

anonymous · Answer

@satellite73 how do i find a and b

anonymous · Answer

never mind! but can u help with c?

anonymous · Answer

you already found b

anonymous · Answer

and i wrote out the answer to c too

anonymous · Answer

i got this for c so far P(x=10) = (200) 140 (-13) 10

anonymous · Answer

i dont get what to do @satellite73

anonymous · Answer

Would (c) be 1/4^20 @satellite73

jamesj · Answer

Yes, that's the answer for c.

anonymous · Answer

so its 9.094?

jamesj · Answer

are you asking if 
$$ \frac{1}{4^{20}} = 9.094\% $$

anonymous · Answer

(1/4)^20 yeah

jamesj · Answer

Definitely not. Calculate again. 4^20 is a huge, huge number.

jamesj · Answer

...and hence 1/4^20 is tiny

anonymous · Answer

I did, it wont give me an answer, can u help

jamesj · Answer

1/4^20 = 9.094947017729282379150390625 × 10^-13

anonymous · Answer

ok so thats the answeR?

jamesj · Answer

Yes, if you are expressing probabilities as decimals.

anonymous · Answer

ok, how would i do d

anonymous · Answer

I have (a)  X is the number of correct guesses
 --> parameter (20,0.25)

--> N= 20 <---

(b) 20 * 1/4 

 = 5

(c)(1/4)^20 
=  9.094947017729282379150390625 × 10^(-13)

(d)  P(x=k)=(nk)pk(1−p)n−k

n=20,
k=10,
p=14

P(x=10) =

jamesj · Answer

yes. So calculate it out

anonymous · Answer

i did and what i got didnt make sense.

jamesj · Answer

What do you get when you do the calculation?

anonymous · Answer

P(x=k)=(nk)pk(1−p)n−k
P(x=10) = (200) 140(1-14)10

anonymous · Answer

i got that..

jamesj · Answer

$$ P(n=10) = {{20}\choose{10}} p^{10}(1-p)^{10} $$ where p = 0.25

anonymous · Answer

where did u get p as that

anonymous · Answer

again, it wont let me calculate that out, can u help

jamesj · Answer

First thing I said: 
"Let p for correct be equal to the probability of guessing one card correctly. Then p = 1/4.  The probability of getting one card wrong is 1 - p = 1 - 1/4 = 3/4."

What's "it"? You definitely need to be able to make these calculations yourself. I can tell you the answer now, but that really won't help you. Do the calculation yourself, so you know you can replicate it yourself, least of all in your tests!

anonymous · Answer

wait what?

anonymous · Answer

this is for d right?

anonymous · Answer

@JamesJ

anonymous · Answer

hellloo

anonymous · Answer

i only  need help with calculating  d.

anonymous · Answer

hi, i think maybe this formula is a mystery, but it should not be
to ask "what is the probability he gets 10 out of 20?" means 
10 are right, 10 are wrong

now if he got 10 right in a row and 10 wrong in a row it would be 
$$\left(\frac{1}{4}\right)^{10}\left(\frac{3}{4}\right)^{10}$$

anonymous · Answer

but there are $\dbinom{20}{10}$ sequences possible of right and wrong, so your answer is 
$$\dbinom{20}{10}\left(\frac{1}{4}\right)^{10}\left(\frac{3}{4}\right)^{10}$$

anonymous · Answer

clearly you need a calculator if you want to get a decimal out of this i would use this http://www.wolframalpha.com/input/?i=%2820+choose+10+%29%281%2F4%29^10%283%2F4%29^10

anonymous · Answer

so the answer comes from

anonymous · Answer

(2010)(14)10(34)10? @satellite73

anonymous · Answer

Is it 0.0099?

anonymous · Answer

yes

anonymous · Answer

thx