# Thread: Independent events problem

1. ## Independent events problem

I'm familiar with how to answer a question about independent events when it asks for "at least one" event occurring, but this one has me stumped, I've looked online everywhere:

Cars coming to a dead end intersection can turn either left or right. Suppose that successive cars choose a turning direction independently of one another and that P(left turn)=0.7. Among the next 3 cars, what is the probability that at least 2 turn left?

I'm not sure where to start (sample spaces?) or where I can use the given probability value (Independent events formula?)

Any help would be appreciated. Thanks in advance.

2. Originally Posted by entropyyy
I'm familiar with how to answer a question about independent events when it asks for "at least one" event occurring, but this one has me stumped, I've looked online everywhere:

Cars coming to a dead end intersection can turn either left or right. Suppose that successive cars choose a turning direction independently of one another and that P(left turn)=0.7. Among the next 3 cars, what is the probability that at least 2 turn left?

I'm not sure where to start (sample spaces?) or where I can use the given probability value (Independent events formula?)

Any help would be appreciated. Thanks in advance.
Dear entropyyy,

First you have to define the events,

First car turning left: A

Second car turning left: B

Third car turning left: C

Third car turning right: D

Now we have to find the event where at least two cars turn left. Using the above abbrevations can you write an expression for this event?

3. Originally Posted by Sudharaka
Dear entropyyy,

First you have to define the events,

First car turning left: A

Second car turning left: B

Third car turning left: C

Third car turning right: D

Now we have to find the event where at least two cars turn left. Using the above abbrevations can you write an expression for this event?
Since they're all independent, it would be
P(A intersect B intersect D) = P(A) x P(B) x P(D)
... right?
And because the only other choice besides turning left is turning right,
P(right turn)=0.3

So the answer is 0.7 x 0.7 x 0.3 ?

4. Originally Posted by entropyyy
Since they're all independent, it would be
P(A intersect B intersect D) = P(A) x P(B) x P(D)
... right?
And because the only other choice besides turning left is turning right,
P(right turn)=0.3

So the answer is 0.7 x 0.7 x 0.3 ?
Dear entropyyy,

You have taken the probability of two cars turning left and the other turning right. But do you realize that the problem says "....at least two cars..." So can you tell me what this means?

5. Originally Posted by Sudharaka
Dear entropyyy,

You have taken the probability of two cars turning left and the other turning right. But do you realize that the problem says "....at least two cars..." So can you tell me what this means?
Right! What I did would be the probability of exactly two cars turning left;
do I need to now add the probability of all three turning left (because all three does include at least two)?

So (0.7 x 0.7 x 0.3) x (0.7)^3 ?

6. Originally Posted by entropyyy
Right! What I did would be the probability of exactly two cars turning left;
do I need to now add the probability of all three turning left (because all three does include at least two)?

So (0.7 x 0.7 x 0.3) x (0.7)^3 ?
Dear entropyyy,

Your idea is correct but your expression is incorrect. You have to add the probability of all three turning left.

That is, Probability that at least two cars turning left= $(0.7 \times 0.7 \times 0.3) + (0.7)^3$

Although the answer is obtained I would like you to show how the problem could be solved in a more methodical way.

Using our abbrevations,

Event that at least two cars turn left= $(A\cap{B}\cap{C})\cup{(A\cap{B}\cap{D}})$

Since the two events $A\cap{B}\cap{C}~and~A\cap{B}\cap{D}$ are mutually exclusive,

Probability that at least two cars turn left= $P(A\cap{B}\cap{C}) + P(A\cap{B}\cap{D})$

Since A,B,C and A,B,D are independent,

Probability that at least two cars turn left= $(P(A)\times{P(B)}\times{P(C)}) +( P(A)\times{P(B)}\times{P(D)})$

Probability that at least two cars turn left= $(0.7\times0.7\times0.7)+(0.7\times0.7\times0.3)$

7. Originally Posted by Sudharaka
Dear entropyyy,

Your idea is correct but your expression is incorrect. You have to add the probability of all three turning left.

That is, Probability that at least two cars turning left= $(0.7 \times 0.7 \times 0.3) + (0.7)^3$

Although the answer is obtained I would like you to show how the problem could be solved in a more methodical way.

Using our abbrevations,

Event that at least two cars turn left= $(A\cap{B}\cap{C})\cup{(A\cap{B}\cap{D}})}$

Since the two events $A\cap{B}\cap{C}~and~A\cap{B}\cap{D}$ are mutually exclusive,

Probability that at least two cars turn left= $P(A\cap{B}\cap{C}) + P(A\cap{B}\cap{D})$

Since A,B,C and A,B,D are independent,

Probability that at least two cars turn left= $P(A)\timesP(B)\timesP(C) + P(A)\timesP(B)\times(D)$

Probability that at least two cars turn left= $(0.7\times0.7\times0.7)+(0.7\times0.7\times0.3)$

Sorry, I meant to put a + sign there, I even wrote it in the sentence before!

This definitely does clear things up: so the probability of at least two left turns would be the sum of the probability of exactly two left turns, and exactly three left turns...
We used the property of independent events to calculate the probability of each event occuring using multiplication...
Then we used the fact that {two left turns} and {three left turns} are mutually exclusive or disjoint, and this justifies adding the probabilities of each event to get the final answer.

Thank you for your thorough and patient explanation, Sudharaka!

8. Originally Posted by entropyyy
Sorry, I meant to put a + sign there, I even wrote it in the sentence before!

This definitely does clear things up: so the probability of at least two left turns would be the sum of the probability of exactly two left turns, and exactly three left turns...
We used the property of independent events to calculate the probability of each event occuring using multiplication...
Then we used the fact that {two left turns} and {three left turns} are mutually exclusive or disjoint, and this justifies adding the probabilities of each event to get the final answer.

Thank you for your thorough and patient explanation, Sudharaka!
Dear entropyyy,

You are always welcome! I cleared all the Latex errors. So you can refer the previous post with ease.

Hope your problems are cleared.