Question

You spin a spinner, and the probability that the spinner will land on red is 18%. Suppose you spin the spinner 4 times. What is the probability that the spinner will land on red at least 2 times? Round to the nearest tenth, if necessary.

10.1%
15.1%
24.3%
63.8%

Answer 1

@whpalmer4 Any good at probability?

Answer 2

Are those the only answer choices? Because I think my process is correct but I'm not getting any of those choices. Would you like me to type up what I did?

Answer 3

@luminate78 yes this is all they give me. The only probability I can do is say that 99% of the questions are this confusing.

Answer 4

Okay see if you can figure out where I'm going wrong

Answer 5

type down your solution

Answer 6

I don't know how to do it. I tried multiplying 9/50 four times but that didn't work... I'm completely lost.

Answer 7

Wait I figured it out

Answer 8

I used complementary probability so 1- P(Red<2 times).

P(Red <2 times) = P(Red = 0) + P(R = 1)

Using Binomial Expansion,
P(Red = 0) = 1(.18)^0 * (.82)^4 = 0.4521276
P(Red = 1) = 4(.18)(.82)^3 = 0.39698496

P(Red> = 2) = 1-(0.39698496+ 0.4521276) = .151

Answer 9

So that's the second answer choice

Answer 10

I don't know what you did.

Answer 11

hmmm? well you're looking for P(Red>= 2) right?

Answer 12

yes

Answer 13

Complements?

Answer 14

wait do you know what complementary probability is?

Answer 15

If I learned it I don't remember it.

Answer 16

Complementary Events

Two events are said to be complementary when one event occurs if and only if the other does not. The probabilities of two complimentary events add up to 1 .

For example, rolling a 5 or greater and rolling a 4 or less on a die are complementary events, because a roll is 5 or greater if and only if it is not 4 or less. The probability of rolling a 5 or greater is = , and the probability of rolling a 4 or less is = . Thus, the total of their probabilities is + = = 1 .

Answer 17

Oh that sounds sort of familiar.

Answer 18

So do you see how I got this: P(Red>= 2 times) = 1- P(Red<2 times).

Answer 19

Not really. Sorry I'm usually really good at math, but I'm lost when it comes to probability.

Answer 20

No it's fine...okay um well you're looking for the probability that the spinner lands on red at least 2 times right? Well sometimes it's easier to find the probability of the complement of an event (see definition posted earlier) than it is to find the probability of the actual event. The probability of two complementary events add up to 1.

The complement of getting at least 2 reds would be getting less than 2 reds (ex. getting no reds or 1 red).

Therefore, we can say that

P(at least 2 reds) + P(getting less than 2 reds) = 1
P(at least 2 reds) = 1- P(getting less than 2 reds)

Answer 21

@rosedewittbukater does that make sense?

Answer 22

Yeah I think so.

Answer 23

Okay so now you're trying to find numerical values for the following equation
P(at least 2 reds) = 1- P(getting less than 2 reds)

Well, the number of times you can land on red is a discrete quantity meaning that you can only get whole number answers. (All this means is that you can land on red 3 times, 4 times, 5 times but not 3.45 times because that wouldn't make any sense)

Does that make sense?

Answer 24

@rosedewittbukater Well since the number of times you can land on red is a discrete quantity, P(landing on red less than 2 times) is only satisfied by P(landing on red once) + P(never landing on red). There are no other options

Answer 25

ok

Answer 26

Okay now do you know how to find P(landing on red once)?

Answer 27

Thats 18% right? so 1 - 82%?

Answer 28

Well um no not quite...I'm asking you to find the probability that after spinning the spinner 4 times, the spinner only landed on red once.

18% is the probability that the spinner will land on red on any given individual spin

Answer 29

So, out of the 4 times your spun the spinner, the spinner only landed on red once

Answer 30

How do you find it for four spins?

Answer 31

Hmm? Are you familiar with binomial probability?

Answer 32

I learned it but don't remember. I just looked in the book and found the formula though.

Answer 33

If S represents success (you land on red) and F represent failure (you don't land on red)

For four spins we can write

SFFF
FSFF
FFSF
FFFS

We can find the probability of one of the events (it's the same for the others)

4*(S)*(F)^3

Do you see how I got this far?

Answer 34

i think so. Does the order of the FS matter?

Answer 35

The formula in my textbook says P(x) = nCxP^xQ^(n-X)

Answer 36

Well if you know the formula then we have the probability of landing on red once during four spins is

(4C1)*(.18)^1 (.82)^3

Answer 37

got that?

Answer 38

Why is x 1?

Answer 39

because you're trying to find P(x=1) or the probability of landing on red only once during your 4 spins

Answer 40

I thought it was of at least twice?

Answer 41

noooo remember

P(at least 2 reds) = 1- P(getting less than 2 reds)

P(at least 2 reds) = 1- (P(landing on red once) + P(never landing on red))

Right now I'm trying to show you how to find P(landing on red once)

Answer 42

ok....

Answer 43

Using the formula, P(landing on red once) = 4(.18)(.82)^3 = 0.39698496.

Now you can use the same formula to find P(never landing on red)

Answer 44

I tried using the formula by myself
(to see if I could try because I feel bad making you do all the work) but I don't know if I did it right. Can I show you what I did and correct me?

Answer 45

yeah

Answer 46

If you don't want to use formulas ( I think they're tedious to memorize), remember P(never landing on red) is equivalent to finding the probability of FFFF

FFFF = 0.82*0.82*0.82*0.82

Answer 47

If you use the formula
P(x = 0) = (4 C 0)*(.18)^0 *(.82)^(4-0)
P(x = 0) = (.82)^4 = 0.4521276

Answer 48

I used the formula.
\[P(x) = _{n}C _{x}P ^{x}q ^{n-x}\]\[P(2) = _{4}C _{2}P ^{2}q ^{4-2}\]\[P(2) = 6(0.18) ^{2}(0.82) ^{2}\]\[P(2) = 6(0.0324)(06724)\] P(2) = 13.07%

Answer 49

It told me to find q the same way you did. 1-probability of not landing on red, which is 1-0.18=0.82.

Answer 50

So Final Answer

P(at least 2 reds) = 1- P(getting less than 2 reds)

P(at least 2 reds) = 1- (P(landing on red once) + P(never landing on red))

P(at least 2 reds) = 1- P(getting less than 2 reds)

P(at least 2 reds) = 1- ( 4(.18)(.82)^3 + (.82)^4 )

P(at least 2 reds) = 1-(0.39698496+ 0.4521276) = .151

Answer 51

Nooooo what you found using the formula is P(Landing on Red exactly twice) the problem is asking for P(landing on red AT LEAST 2 times)

Answer 52

Ooohh ok. I'm still confused but I guess it makes sense because 15.1% is a little bigger than 13%

Answer 53

This is the entire process at once:

P(at least 2 reds) = 1- P(getting less than 2 reds)

P(at least 2 reds) = 1- (P(landing on red once) + P(never landing on red))

P(at least 2 reds) = 1- ( 4(.18)(.82)^3 + (.82)^4 )

P(at least 2 reds) = 1-(0.39698496+ 0.4521276) = .151

Answer 54

Ok I think I understand what you did. I don't know why adding the prob of landing once + not landing = getting less than 2, but I think I get the rest.

Answer 55

hmmm? well what other ways can you get less than 2 when you're spinning a spinner?

Answer 56

I don't really know haha.

Answer 57

You can't land on red 1.8 times or 0.345 times. It doesn't make any sense. The number of times you land on red has to be a whole number quantity

Answer 58

The only whole number less than 2 are 0 and 1

Answer 59

Ok...

Answer 60

So the answer is 15.1% right?

Answer 61

So in this case, P(x<2) = P(x=0)+P(x =1)

I think so

Answer 62

Ok. Sorry it was so hard to explain to me. Probability just isn't my thing. But thank you sooo much for your help and effort. I really appreciate it.

Answer 63

Yeah of course