Bayes' Theorem and a car starting probability

[CivilSigma]

Homework Statement

A man owns two old cars, A and B, and has trouble starting them on cold mornings. The probability both will start is 0.1; the probability B starts and A does not is 0.1; the probability that neither starts is 0.4

a) Find the probability that car A will start.

b) Find the probability that car A will start, given car B starts.

c) Find the probability that car B will start, given car A starts.

Homework Equations

P(A|B) = P(A and B)/P(B)

The Attempt at a Solution

From the question:

P(A and B) = 0.1
P(B|A') = 0.1 ----> Which implies that P(A and B) + P(B|A) = P(B)
P( B' and A') = 0.4 which implies P(A and B) = 0.6

I have drew the Venn Diagram, and I concluded the following:

P(B) = 0.2
P(A)= 0.6

But I am having a hard time deriving them using the equation of Bayes Theorem the general mathematical approach

So, to answer

a) P(A)=0.6
b) P(A|B) = P(A and B)/ P(B) = 0.1/0.2 = 0.5
c) P(B|A) = P(B and A)/P(A) = 0.1/0.6 = 0.166

Can some one please explain to me how you obtain the solution using equations?

 

[kuruman]

First things first. If the man tries to start the cars one afer the other there are four outcomes for starting: Both, only A, only B and none. Their sum must be 1. Is the sum 1 in part (a)?

 

[CivilSigma]

First things first. If the man tries to start the cars one afer the other there are four outcomes for starting: Both, only A, only B and none. Their sum must be 1. Is the sum 1 in part (a)?

Are all the problem statements conditional probabilities if the owner starts car A then B?

So we would have:

P( A | B) = 0.1
P (A | B') = (Not given)
P (B |A') = 0.1
P (B'|A') = 0.4

Which means the probability of car A starting and B not starting is 0.4.

I also know that : P(A) = P(A|B) P(B) + P(A|B')P(B')

 

[Ray Vickson]

Homework Statement

A man owns two old cars, A and B, and has trouble starting them on cold mornings. The probability both will start is 0.1; the probability B starts and A does not is 0.1; the probability that neither starts is 0.4

a) Find the probability that car A will start.

b) Find the probability that car A will start, given car B starts.

c) Find the probability that car B will start, given car A starts.

Homework Equations

P(A|B) = P(A and B)/P(B)

The Attempt at a Solution

From the question:

P(A and B) = 0.1
P(B|A') = 0.1 ----> Which implies that P(A and B) + P(B|A) = P(B)
P( B' and A') = 0.4 which implies P(A and B) = 0.6

I have drew the Venn Diagram, and I concluded the following:

P(B) = 0.2
P(A)= 0.6

But I am having a hard time deriving them using the equation of Bayes Theorem the general mathematical approach

So, to answer

a) P(A)=0.6
b) P(A|B) = P(A and B)/ P(B) = 0.1/0.2 = 0.5
c) P(B|A) = P(B and A)/P(A) = 0.1/0.6 = 0.166

Can some one please explain to me how you obtain the solution using equations?

Part (a) can be done without using Bayes: you just have three events and their complements, together with some given information:
$$ \begin{array}{cccl}
P(A \cap B)&=& 0.1 &\text{(both A and B start)} \\
P(B \cap \bar{A}) &=& 0.1& \text{(B starts and A does not)}\\
P(\bar{A} \cap \bar{B}) &=& 0.4 & \text{(neither starts)}
\end{array}
$$ (Here, $\bar{E}$ denotes the complement of an event $E$.)

It might be easiest to obtain $P(A)$ as $1-P(\bar{A})$ and then figure out how to get $P(\bar{A}).$

I get $P(A) = 0.5$ and $P(B) = 0.2.$

 

[CivilSigma]

This makes way more sense @Ray C

So,

$$P(A)=1-P(A')$$
$$P(A') = P(B)-P(A and B) + P( A' and B') = P(B) -0.1+0.4$$
$$\therefore P(A)=1-P(B)-0.3$$

We also know:

$$P(A and B)=0.1$$

Since the events are independent:

$$P(A and B) = P(A)(B)$$

Now substitute in P(A)

$$0.1=(1-P(B)-0.3 \cdot P(B) = 0.1$$

This gives us a quadratic in P(B) and when I solve it I get:

$$P(B) = 0.5 or 0.2$$

How do you determine which one of these is extraneous? They are both valid since they are below 1

Edit: I know, I check the third given probability:

$$P(B \, A^C) = 0.1$$
This only works when P(A) = 0.5Thank you so much :)

 

[Ray Vickson]

This makes way more sense @Ray C

So,

$$P(A)=1-P(A')$$
$$P(A') = P(B)-P(A and B) + P( A' and B') = P(B) -0.1+0.4$$
$$\therefore P(A)=1-P(B)-0.3$$

We also know:

$$P(A and B)=0.1$$

Since the events are independent:

$$P(A and B) = P(A)(B)$$

Now substitute in P(A)

$$0.1=(1-P(B)-0.3 \cdot P(B) = 0.1$$

This gives us a quadratic in P(B) and when I solve it I get:

$$P(B) = 0.5 or 0.2$$

How do you determine which one of these is extraneous? They are both valid since they are below 1

Edit: I know, I check the third given probability:

$$P(B \, A^C) = 0.1$$
This only works when P(A) = 0.5Thank you so much :)

The events $B \cap \bar{A}$ and $B \cap A$ are disjoint (that is, mutually exclusive) and their union is just $B$ itself. The addition law for disjoint events gives $$P(B) = P(B \cap \bar{A}) + P(B \cap A) = 0.1 + 0.1 = 0.2$$ No quadratic equation is needed, and there is no ambiguity about the correct value.