Driving home poem
which tell us that the poet is falling asleep
Answers
Answer:
Probability = 6/23
Step-by-step explanation:
Given:
The probability of X,Y, Z becoming managers are 4/9, 2/9 and 1/3 respectively
The probability that bonus scheme will be introduced if X, Y, Z become managers are 3/10, 1/2 and 4/5 respectively.
To Find:
The probability that the manager appointed was X if the bonus scheme has been introduced.
Solution:
Let B₁, B₂, B₃ be the events of X, Y, Z becoming managers respectively.
Hence by given,
\sf P(B_1)=\dfrac{4}{9} ,\: \: P(B_2)=\dfrac{2}{9},\: \: P(B_3)=\dfrac{1}{3}P(B
1
)=
9
4
,P(B
2
)=
9
2
,P(B
3
)=
3
1
Now let A be the event that the bonus scheme is introduced.
Hence,
\sf P(A/B_1)=\dfrac{3}{10}P(A/B
1
)=
10
3
\sf P(A/B_2)=\dfrac{1}{2}P(A/B
2
)=
2
1
\sf P(A/B_3)=\dfrac{4}{5}P(A/B
3
)=
5
4
Here we have to find the probability that X is the manager if the bonus scheme has been introduced, ie, P(B₁/A)
By Bayes Theorem,
\sf P(B_1/A)=\dfrac{P(B_1)\times P(A/B_1)}{P(B_1)\times P(A/B_1)+P(B_2)\times P(A/B_2)+P(B_3)\times P(A/B_3)}P(B
1
/A)=
P(B
1
)×P(A/B
1
)+P(B
2
)×P(A/B
2
)+P(B
3
)×P(A/B
3
)
P(B
1
)×P(A/B
1
)
Substitute the data,
\sf P(B_1/A)=\dfrac{4/9\times 3/10}{4/9\times 3/10+2/9\times 1/2+1/3\times 4/5}P(B
1
/A)=
4/9×3/10+2/9×1/2+1/3×4/5
4/9×3/10
\sf P(B_1/A)=\dfrac{2}{15} \div\bigg(\dfrac{2}{15} +\dfrac{1}{9} +\dfrac{4}{15}\bigg)P(B
1
/A)=
15
2
÷(
15
2
+
9
1
+
15
4
)
\sf P(B_1/A)=\dfrac{2}{15} \div\bigg(\dfrac{2}{5} +\dfrac{1}{9} \bigg)P(B
1
/A)=
15
2
÷(
5
2
+
9
1
)
\sf P(B_1/A)=\dfrac{2}{15} \div\bigg(\dfrac{18+5}{45} \bigg)P(B
1
/A)=
15
2
÷(
45
18+5
)
\sf P(B_1/A)=\dfrac{2}{15} \times\bigg(\dfrac{45}{23} \bigg)P(B
1
/A)=
15
2
×(
23
45
)
\sf P(B_1/A)=\dfrac{6}{23}P(B
1
/A)=
23
6
Hence the probability is 6/23.