Question

Dilbert forgot to study for the multiple-choice exam. Each question has four choices and there are 50 questions. Dilbert, in his infinite wisdom, decides to randomly guess on each problem. Find the probability for each:
a) If Dilbert gets 30 or more right, he will be named the company CEO.
b) If Dilbert gets 20 to 29 right, he will be promoted and get a 10% raise.
c) If Dilbert gets 10 to 19 right, he will get to keep his menial job and pay.
d) If Dilbert gets less than 10 right, he will be kicked out of cubicle central and unemployed.

Answer 1

@jim_thompson5910 ?

Answer 2

this is a pain to do by hand, but luckily there's a calculator especially designed for this sort of thing

Answer 3

ok!

Answer 4

how would i do it ?

Answer 5

here is the calculator

http://stattrek.com/online-calculator/binomial.aspx

Answer 6

this calculator uses what is known as the binomial distribution

Answer 7

does it look easy to use?

Answer 8

hmmmm

Answer 9

sort of?

Answer 10

Probability of success on a single trial
Number of trials
Number of successes (x)
Binomial probability: P(X = x)
Cumulative probability: P(X < x)
Cumulative probability: P(X < x)
Cumulative probability: P(X > x)
Cumulative probability: P(X > x)

Answer 11

probability of success = 1/4 = 0.25 since the chances of him getting a single question right is 1 out of 4

Answer 12

so that's your first box

Answer 13

k i put that

Answer 14

number of trials is 50

basically you have 50 questions or 50 events where a guess is taking place

Answer 15

ok thats what i thought!!

Answer 16

for this problem, the first two boxes will remain the same

0.25

and

50

Answer 17

the third box will vary as you answer parts a) through d)

Answer 18

ok so would the third box be 30 for a?

Answer 19

parts b) and c) are a bit more complicated and will have to be broken down into parts

Answer 20

yes

Answer 21

my answer came outas 1.29633040882142E-07

Answer 22

for binomial prob.... can u try it?

Answer 23

that's the probability if you wanted exactly 30 questions correct

but you want 30 or more

Answer 24

X = 30 ... exactly 30

X >= 30 ... 30 or more

Answer 25

so how would i do that?

Answer 26

so its 3.45955140090837E-08

Answer 27

its the last box, it should be there

Answer 28

see the X >= 30

Answer 29

1.64?

Answer 30

you're seeing X > 30, which is different

Answer 31

yeah 1.64 x 10^(-7)

Answer 32

in any event, all of these probabilities are very small, practically zero

Answer 33

but it's probably best to be as accurate as possible

Answer 34

1.64228555 then?

Answer 35

times 10^(-7)

yeah

Answer 36

or 1.64 x 10^(-7)

Answer 37

you can probably get away with rounding, depends on the book

Answer 38

the second one is probably easier to handle

Answer 39

ok i put that

Answer 40

is b 1/4? i had gotten b a while ago, im not sure if its right tho

Answer 41

no 1/4 doesn't sound right (it sounds too big), but let me check

Answer 42

kk

Answer 43

yeah way too big, it should be 0.88% or 0.0088

you basically compute both P(X <= 29) and P(X < 20) and you subtract the two probabilities to find P(20 <= X <= 29)

Answer 44

so that's what I meant when I said you had to break up parts b) and c) into smaller pieces

Answer 45

so b is .0088

Answer 46

approximately, yes

Answer 47

can u help with c? sorry

Answer 48

so right of the bat, we see that he has a better chance of getting 20 to 29 questions correct than getting 30 or more questions correct

Answer 49

granted 0.88% isn't that big a chance, but it's much larger than 1.64 x 10^(-7)

Answer 50

yea

Answer 51

can you tell me what P(X < 20) is?

Answer 52

for what value?

Answer 53

first two boxes are the same

but change the third box to 20

that will show you P(X < 20)

Answer 54

its .986

Answer 55

not X <= 20

X < 20

Answer 56

yea its 0.986082391321338

Answer 57

it should be roughly 0.99119684178598

so about 0.9911

Answer 58

you must be looking at the wrong box

Answer 59

the box says that

Answer 60

hmm one sec

Answer 61

oh my bad, i typed in the wrong number lol

Answer 62

yeah it's 0.98608 roughly

Answer 63

now do the same for P(X < 10)

Answer 64

ok i got that then d is .1636??

Answer 65

so is this right:

a) 1.64 x 10^(-7)

b) 0.0088

c) 0.98608

d).1636

Answer 66

P(X < 10) = 0.1636839

now subtract the two probabilities

0.98608 - 0.1636839

0.8223961

So P(10 <= X <= 19) = 0.8223961

Answer 67

a) 1.64 x 10^(-7)

b) 0.0088

c) 0.98608

d)0.8223961

Answer 68

So if you round to 4 places, it's roughly 0.8224

Answer 69

are these the right answers?

Answer 70

no we haven't even started part d) yet

Answer 71

you're moving too fast lol

Answer 72

oh ok 0.8223961 is c then

Answer 73

sorry haha

Answer 74

yes roughly

Answer 75

a) 1.64 x 10^(-7)

b) 0.0088

c) 0.8223961

d)

Answer 76

so r these right haha

Answer 77

luckily, we have the answer to part d) because we found P(X < 10) in part c

Answer 78

what was P(X < 10) again

Answer 79

im not sure

Answer 80

0.163683900250091
?

Answer 81

I wrote it above in the steps to get the answer to part c

Answer 82

So, c is 0.8224?

Answer 83

yes after rounding

Answer 84

and yes part d is roughly 0.16368

Answer 85

ok so d?

Answer 86

a) 1.64 x 10^(-7)

b) 0.0088

c) 0.8224

d)

Answer 87

@jim_thompson5910 how do i find d?

Answer 88

i just said it

Answer 89

would d be 0.163683900250091 since thats x<10?

Answer 90

oh o ksoryy lol

Answer 91

yes since P(X < 10) = 0.16368 which is what we found in part c

Answer 92

a) 1.64 x 10^(-7)

b) 0.0088

c) 0.8224

d) 0.16368

Answer 93

is that right

Answer 94

thanks for ur help @jim_thompson5910

Answer 95

yes all four look great

Answer 96

I would keep practicing with that calculator as well

if you have a TI-83, you should be able to do the same things

Answer 97

or a TI-86, 89, etc