Question

One bag contains 4 white and 5 black balls. Another bag contains 6 white and 7 black balls. A ball is transferred from the first bag to the second bag and then a ball is drawn from the second bag. If the probability that the ball drawn is white $\dfrac{{29}}{m}$ , find $m$.

Answer 1

Probability is a measure of the likeliness of an event to occur. Most of the events cannot be predicted with total certainty.

P\left( A \right) = \left( {P\left( {{E_1}} \right) \times P\left( {A/{E_1}} \right)} \right) + \left( {P\left( {{E_2}} \right) \times P\left( {A/{E_2}} \right)} \right)$$ So by using this concept we can solve the above given question. **Complete step by step solution:** Given bag contains: ${\text{Bag}}\;1: \\\ \Rightarrow 4\;{\text{white balls}} \\\ \Rightarrow 5\;{\text{black balls}} \\\ {\text{Bag}}\;2: \\\ \Rightarrow 6\;{\text{white balls}} \\\ \Rightarrow 7\;{\text{black balls}} \\\ {\text{Probability that the ball drawn is white:}}\dfrac{{29}}{m}...................................\left( i \right) \\\ $ Now we need to find $m$. Also we know that a ball was transferred from the first bag to the second bag. Also let: $${E_1} = {\text{A black ball is transferred from first to second bag}}{\text{.}} \\\ \Rightarrow{E_2} = {\text{A white ball is transferred from first to second bag}}{\text{.}} \\\ \Rightarrow A = {\text{A white ball is drawn}}{\text{.}} \\\ $$ Now let’s take two different cases and thus find the corresponding probability: 1st case: The transferred ball from Bag 1 is black: We also know that: ${\text{Bag}}\;1: \\\ \Rightarrow 4\;{\text{white balls}} \\\ \Rightarrow 5\;{\text{black balls}} \\\ $ So we can write: $P\left( {{E_1}} \right) = \dfrac{5}{9}...........................................\left( {ii} \right)$ 2nd case: The transferred ball from Bag 1 is white: We also know that: ${\text{Bag}}\;1: \\\ \Rightarrow 4\;{\text{white balls}} \\\ \Rightarrow 5\;{\text{black balls}} \\\ $ So we can write: $P\left( {{E_2}} \right) = \dfrac{4}{9}...........................................\left( {iii} \right)$ Now we know that: ${\text{Bag}}\;2: \\\ \Rightarrow 6\;{\text{white balls}} \\\ \Rightarrow 7\;{\text{black balls}} \\\ $ Here the total number of balls would be 14 since a ball is transferred from Bag 1 to Bag 2. Such that: $P\left( {A/{E_1}} \right) = \dfrac{6}{{14}}...........................................\left( {iv} \right) \\\ P\left( {A/{E_2}} \right) = \dfrac{7}{{14}}...........................................\left( v \right) \\\ $ Now from the properties of probability we can write: $$P\left( A \right) = \left( {P\left( {{E_1}} \right) \times P\left( {A/{E_1}} \right)} \right) + \left( {P\left( {{E_2}} \right) \times P\left( {A/{E_2}} \right)} \right)$$ Such that: $$P\left( A \right) = \left( {\dfrac{5}{9} \times \dfrac{6}{{14}}} \right) + \left( {\dfrac{4}{9} \times \dfrac{7}{{14}}} \right) \\\ \Rightarrow P\left( A \right)= \left( {\dfrac{{30}}{{126}}} \right) + \left( {\dfrac{{28}}{{126}}} \right) \\\ \Rightarrow P\left( A \right)= \dfrac{{58}}{{126}} \\\ \Rightarrow P\left( A \right)= \dfrac{{29 \times 2}}{{63 \times 2}} \\\ \therefore P\left( A \right)= \dfrac{{29}}{{63}}................................\left( {vi} \right) \\\ $$ Now we know that $P\left( A \right)$ simply means ${\text{Probability that the ball drawn is white}}$. Now it’s already given in the question as: $\dfrac{{29}}{m}$. Such that on comparing to equation (vi) we can write: $m = 63$ **Therefore our final answer is $m = 63$.** **Note:** One can predict only the chance of an event to occur which simply means how likely they are to happen, using it. Probability can range in from 0 to 1, where 0 means the event to be an impossible one and 1 indicates a certain event. Also one should be thorough with the properties of probabilities for solving questions similar to these.