Probability

Question

A slot machine has 3 wheels. Each wheel has 10 positions: a bar, and digits 0,1,2... When the handle is pulled the three wheels spin independently before coming to rest. Find the probability of:

 

• Getting all three bars.
• Getting the same number on each wheel.
• Getting at least one bar.

 

 

Answer 1

Hi Maddie,

Probability always seems a little trickier when they make up these odd scenarios. Let's start by translating the problem into probability terms:

 

We have:

3 independent events (each wheel is considered an event, the wheels don't interact

10 possible outcomes for each (each wheel position is a potential outcome)

 

1) Conditional probability with independent events --> we multiply each individual probability

 

P(bar) * P(bar) * P(bar)

Each P(bar) = .1 <-- 1 out of 10 possible outcomes

 

2) This is a conditional probability where the first event is independent and the next two depend on the result of the first.

 

P(any number) * P(# on wheel 1) * P(# on wheel 1)

Probability of getting any number = .9 <-- 9 numbers out of 10

Each P(# on wheel 1) = .1   <-- 1 out of 10, we want a specific number

 

3) Since the sum of all possible outcomes is 1, we have an easy way to solve for getting "at least 1 bar."

P(at least 1 bar) = 1 - P(no bars)

= 1 - .9*.9*.9 <-- no bars are each 9 out of 10

Answer 2

Probability for 1 bar on 1 wheel                                  1/10

Probability for no bar on 1 wheel         1  - 1/10  =     9/10

 

When you have multiple selections(3 wheels) you would solve a problem like this.

 

 

(probability 1)  * (probability 2) * (probability 3)   = Total probability

 

Think about all 3 bars. You need to get 1 bar on 1 wheel 3 times.

 

(1/10) * (1/10) * (1/10) = 1/1000

 

Think about the same number on each wheel

 

Probability to get ANY number * probability to get SAME number * probability to get SAME number

 

 

The last one is a little harder. You could (METHOD A) find the chance to get 1 bar, 2 bars and 3 bars and add them all up. OR You could (METHOD B) find the chance to get no bars and subtract that from 1.

 

Think about METHOD B, does this make sense? The chance to get at least 1 bar is the opposite of getting no bars.

 

Probability of no bar * Probability of no bar * Probability of no bar = Probability of 3 no bars

1 - (Probability of 3 no bars) = Probability of at least one bar

 

 

 

If you don't have much experience with probability, here is a simpler example:

 

 

 

You have 3 coins. With side H for head and T for tail.

Possible outcomes... HHH , HHT, HTH, THH, TTH, THT, HTT, TTT (8 outcomes)

Chance to get all heads... (1/2)*(1/2)*(1/2) = 1/8 

Check your answer... HHH is one outcome out of 8 outcomes

 

Chance to get at least head... which of the choices have at least one head? (There are 7)

                                              Which of the choices have no heads? (Only TTT)

                                              The chance to get at least one head is     1 - (chance of no heads).

This works the same way with the slot machine problem.

I hope that helped. Let me know if anything else is difficult.