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) =
Answers
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:
The value of .
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.,
The joint probability in terms of conditional probability is given as:
The conditional probability of an event Y provided that another event X is taking place at the same interval of time is expressed as:
For the provided scenario, a sample space is described by two events and .
It is provided that,
Using the concept of total probability, determine the probability of event B as follows:
Now compute the conditional probability of event given another event as follows:
Thus, the value of .