Question

44. The probability of X, Y & Z becoming
managers are 4/9, 2/9& 1/3 respectively.
The probability that the bonus scheme
will be introduced if X, Y & Z becomes
managers are 3/10,1/2 & 4/5respectively.
Then probability that the manager
appointed was X, if the bonus scheme
has been introduced is​

Answer 1

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,

$\sf P(B_1)=\dfrac{4}{9} ,\: \: P(B_2)=\dfrac{2}{9},\: \: P(B_3)=\dfrac{1}{3}$

Now let A be the event that the bonus scheme is introduced.

Hence,

$\sf P(A/B_1)=\dfrac{3}{10}$

$\sf P(A/B_2)=\dfrac{1}{2}$

$\sf P(A/B_3)=\dfrac{4}{5}$

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,

$\sf P(B_1/A)=\dfrac{P(B_1)\times P(A/B_1)}{P(B_1)\times P(A/B_1)+P(B_2)\times P(A/B_2)+P(B_3)\times P(A/B_3)}$

Substitute the data,

$\sf P(B_1/A)=\dfrac{4/9\times 3/10}{4/9\times 3/10+2/9\times 1/2+1/3\times 4/5}$

$\sf P(B_1/A)=\dfrac{2}{15} \div\bigg(\dfrac{2}{15} +\dfrac{1}{9} +\dfrac{4}{15}\bigg)$

$\sf P(B_1/A)=\dfrac{2}{15} \div\bigg(\dfrac{2}{5} +\dfrac{1}{9} \bigg)$

$\sf P(B_1/A)=\dfrac{2}{15} \div\bigg(\dfrac{18+5}{45} \bigg)$

$\sf P(B_1/A)=\dfrac{2}{15} \times\bigg(\dfrac{45}{23} \bigg)$

$\sf P(B_1/A)=\dfrac{6}{23}$

Hence the probability is 6/23.

Answer 2

Answer

P(x)=

9

4

,P(y)=

9

2

,P(z)=

3

1

Let event A be the bonus scheme introduced

so, P(A/x)=

10

3

and P(A/y)=

2

1

(1) P(A)=P(x).P(a/x)+P(y).P(A/y)+P(z).P(A/z)

=

9

4

×

10

3

+∣cfrac29×

2

1

+

2

1

×

5

2

=

45

6+5+6

=

45

17

(2) Probability that manager appointed if bonus introduced

P(z/A)=

P(A)

P(A/z).P(z)

=

15×17

2×45

=

17

6