So you have 3 phone companies: Company A has a 41% share on the market, B has 38% and C has 21%, summing up gives 100%, so they are really the only ones.

Now they say that 35% of A and B and 30% of C company customers are unsatisfied with the service.

What is probability of a satisfied customer be connected to company B?

Can anyone help me solve this, thank you!

2. Originally Posted by Wright
B has 38% ....

Now they say that 35% of B are unsatisfied with the service.

What is probability of a satisfied customer be connected to company B?
I would suggest 35% of 38% will 'stay' connected.

$\displaystyle \frac{35}{100}\times\frac{38}{100}= \dots$

3. Originally Posted by Wright
So you have 3 phone companies: Company A has a 41% share on the market, B has 38% and C has 21%, summing up gives 100%, so they are really the only ones.

Now they say that 35% of A and B and 30% of C company customers are unsatisfied with the service.

What is probability of a satisfied customer be connected to company B?
I interpreted the question to mean: What proportion of all satisfied customers are connected to company B? Since 65% of A and B customers, and 70% of C customers, are satisfied, the answer should then be $\displaystyle \frac{38\times0.65}{41\times0.65 + 38\times0.65 + 21\times0.7}$.

4. Hello, Wright!

I agree with Opalg . . . this is Conditional Probability.

You have 3 phone companies: Company A has a 41% share on the market,
B has 38% and C has 21%, summing up gives 100%, so they are really the only ones.

Now they say that: 35% of A and B and 30% of C company customers are unsatisfied with the service.

What is probability of a satisfied customer be connected to company B?

We have: .$\displaystyle \begin{array}{ccc} P(A) &=& 0.41 \\ P(B) &=& 0.38 \\ P(C) &=& 0.21 \end{array}$

Let: . $\displaystyle \begin{array}{ccc}S &=& \text{satisfied} \\ U &=& \text{unsatisfied} \end{array}$

Then: .$\displaystyle \begin{array}{ccccccc} P(A_U) &=& 0.35 & \Rightarrow & P(A_S) &=& 0.65 \\ P(B_U) &=& 0.35 & \Rightarrow & P(B_S) &=& 0.65 \\ P(C_U) &=& 0.30 & \Rightarrow & P(C_S) &=& 0.70 \end{array}$

Bayes' Theorem: .$\displaystyle P(B\,|\,S) \;=\;\frac{P(B\,\wedge\,S)}{P(S)}\;\;{\color{blue}[1]}$

Numerator: .$\displaystyle P(B\,\wedge S) \:=\:(0.38)(0.65) \:=\:0.247\;\;{\color{blue}[2]}$

Denominator:

. . $\displaystyle \begin{array}{ccccccc}P(A \wedge S) &=& (0.41)(0.65) &=& 0.2665 \\ P(B \wedge S) &=& (0.38)(0.65) &=& 0.2470 \\ P(C \wedge S) &=& (0.21)(0.30) &=& 0.1470 \\ \hline \text{Total:}& & P(S) &= & 0.6605 & {\color{blue}[3]} \end{array}$

Substitute [2] and [3] into [1]: .$\displaystyle P(B\,|\,S) \;=\;\frac{0.247}{0.6605} \;\approx\;37.4\%$

5. Thanks for the kind answers everyone.