Question

The events A1 and A2 form the
partition of the sample space. If
P(A1) = P(A2) = 1/2, P(BIA1) = 2/5,
P(B|A2) = 4/15 then P(A1B) =​

Answer 1

Answer:

The events A_1A

1

and A_2A

2

form the partition of the sample space, which means,

A_1\cup A_2=SA

1

∪A

2

=S

A_1\cap A_2=\phiA

1

∩A

2

=ϕ

That's why,

\longrightarrow P(S)=P(A_1)+P(A_2)=1⟶P(S)=P(A

1

)+P(A

2

)=1

Given,

P(A_1)=P(A_2)=\dfrac{1}{2}P(A

1

)=P(A

2

)=

2

1

P(B\,|\,A_1)=\dfrac{2}{5}P(B∣A

1

)=

5

2

P(B\,|\,A_2)=\dfrac{4}{15}P(B∣A

2

)=

15

4

We need to find P(A_1\,|\,B)P(A

1

∣B)

We see that,

\longrightarrow B=B\cap S⟶B=B∩S

\longrightarrow B=B\cap(A_1\cup A_2)⟶B=B∩(A

1

∪A

2

)

\longrightarrow B=(B\cap A_1)\cup(B\cap A_2)\quad\quad\dots(1)⟶B=(B∩A

1

)∪(B∩A

2

)…(1)

Also,

\longrightarrow(B\cap A_1)\cap(B\cap A_2)=B\cap A_1\cap B\cap A_2⟶(B∩A

1

)∩(B∩A

2

)=B∩A

1

∩B∩A

2

\longrightarrow(B\cap A_1)\cap(B\cap A_2)=B\cap A_1\cap A_2⟶(B∩A

1

)∩(B∩A

2

)=B∩A

1

∩A

2

\longrightarrow(B\cap A_1)\cap(B\cap A_2)=B\cap(A_1\cap A_2)⟶(B∩A

1

)∩(B∩A

2

)=B∩(A

1

∩A

2

)

\longrightarrow(B\cap A_1)\cap(B\cap A_2)=B\cap\phi⟶(B∩A

1

)∩(B∩A

2

)=B∩ϕ

\longrightarrow(B\cap A_1)\cap(B\cap A_2)=\phi\quad\quad\dots(2)⟶(B∩A

1

)∩(B∩A

2

)=ϕ…(2)

(1) and (2) mean the events B\cap A_1B∩A

1

and B\cap A_2B∩A

2

form the partition of B,B, hence,

\longrightarrow P(B)=P(B\cap A_1)+P(B\cap A_2)\quad\quad\dots(3)⟶P(B)=P(B∩A

1

)+P(B∩A

2

)…(3)

But,

P(B\cap A_1)=P(B\,|\,A_1)\cdot P(A_1)P(B∩A

1

)=P(B∣A

1

)⋅P(A

1

)

P(B\cap A_2)=P(B\,|\,A_2)\cdot P(A_2)P(B∩A

2

)=P(B∣A

2

)⋅P(A

2

)

Then (3) becomes,

\longrightarrow P(B)=P(B\,|\,A_1)\cdot P(A_1)+P(B\,|\,A_2)\cdot P(A_2)⟶P(B)=P(B∣A

1

)⋅P(A

1

)+P(B∣A

2

)⋅P(A

2

)

\longrightarrow P(B)=\dfrac{2}{5}\cdot\dfrac{1}{2}+\dfrac{4}{15}\cdot\dfrac{1}{2}⟶P(B)=

5

2

⋅

2

1

+

15

4

⋅

2

1

\longrightarrow P(B)=\dfrac{1}{5}+\dfrac{2}{15}⟶P(B)=

5

1

+

15

2

\longrightarrow P(B)=\dfrac{1}{3}⟶P(B)=

3

1

Then,

\longrightarrow P(A_1\,|\,B)=\dfrac{P(A_1\cap B)}{P(B)}⟶P(A

1

∣B)=

P(B)

P(A

1

∩B)

\longrightarrow P(A_1\,|\,B)=\dfrac{P(B\cap A_1)}{P(B)}⟶P(A

1

∣B)=

P(B)

P(B∩A

1

)

\longrightarrow P(A_1\,|\,B)=\dfrac{P(B\,|\,A_1)\cdot P(A_1)}{P(B)}⟶P(A

1

∣B)=

P(B)

P(B∣A

1

)⋅P(A

1

)

\longrightarrow P(A_1\,|\,B)=\dfrac{\left(\dfrac{2}{5}\cdot\dfrac{1}{2}\right)}{\left(\dfrac{1}{3}\right)}⟶P(A

1

∣B)=

(

3

1

)

(

5

2

⋅

2

1

)

\longrightarrow\underline{\underline{P(A_1\,|\,B)=\dfrac{3}{5}}}⟶

P(A

1

∣B)=

5

3

Hence 3/5 is the answer.

Answer 2

Answer:

The value of $P(A_{1}|B)=\frac{3}{5}$.

Step-by-step explanation:

The law of total probability is stated as follows:

The total probability of X is computed from the joint probability of event X and events of Y and Y', i.e.,

$P(X)=P(X\cap Y)+P(X\cap Y')$

The joint probability in terms of conditional probability is given as:

$P(X\cap Y)=P(Y|X) \cdot P(X)$

The conditional probability of an event Y provided that another event X is taking place at the same interval of time is expressed as:

$P(Y|X)=\frac{P(X|Y) \cdot P(Y)}{P(X)}$

For the provided scenario, a sample space is described by two events $A_{1}$ and $A_{2}$.

It is provided that,

$P(A_{1})=P(A_{2})=\frac{1}{2}$

$P(B|A_{1})=\frac{2}{5}\\\\P{(B|A_{2})=\frac{4}{15}$

Using the concept of total probability, determine the probability of event B as follows:

$P(B)=P(B\cap A_{1})+P(B\cap A_{2})\\\\P(B)=P(B|A_{1})P(A_{1})+P(B|A_{2})P(A_{2})\\\\P(B)=(\frac{2}{5}\times \frac{1}{2})+(\frac{4}{15}\times \frac{1}{2})\\\\P(B)=\frac{1}{5}+\frac{2}{15}\\\\P(B)=\frac{3+2}{15}\\\\P(B)=\frac{5}{15}\\\\P(B)=\frac{1}{3}$

Now compute the conditional probability of event $A_{1}$ given another event $B$ as follows:

$P(A_{1}|B)=\frac{P(B|A_{1}) \cdot P(A_{1})}{P(B)}\\\\P(A_{1}|B)=\frac{\frac{2}{5}\times \frac{1}{2}}{\frac{1}{3}}\\\\P(A_{1}|B)=\frac{1}{5}\times \frac{3}{1}\\\\P(A_{1}|B)=\frac{3}{5}$

Thus, the value of $P(A_{1}|B)=\frac{3}{5}$.