Answer

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**Hint:**

We start solving the problem by finding the probability of choosing balls and drawing white and black balls from each bag. We then find the possible ways of drawing exactly four white balls from the bags using the given condition. We then make use of the obtained probabilities and make the necessary calculations to get the required probability.

**Complete step by step answer:**

According to the problem, we are given that one bag contains 2 white balls and 2 black balls and another contains 3 white balls and 5 black balls. A bag and one ball from that bag is chosen at random and the ball is replaced. This trail is repeated five times. We need to find the probability that there are exactly four white balls drawn.

We can see that there are two bags so, the probability of choosing each bag is $ {{P}_{1}}={{P}_{2}}=\dfrac{1}{2} $.

Now, let us find the probability of choosing a white and black ball from the first bag.

We get the probability of choosing a white ball from first bag as $ {{P}_{W1}}=\dfrac{2}{2+2}=\dfrac{2}{4}=\dfrac{1}{2} $ .

We get the probability of choosing a black ball from first bag as $ {{P}_{B1}}=\dfrac{2}{2+2}=\dfrac{2}{4}=\dfrac{1}{2} $ .

Now, let us find the probability of choosing white and black balls from the second bag.

We get the probability of choosing a white ball from the second bag as $ {{P}_{W2}}=\dfrac{3}{3+5}=\dfrac{3}{8} $.

We get the probability of choosing a black ball from the second bag as $ {{P}_{B2}}=\dfrac{5}{3+5}=\dfrac{5}{8} $ .

Now, we need to find the probability of drawing exactly four white balls if a bag and one ball from that bag is chosen at random and the ball is replaced five times.

So, the possibilities are drawing exactly four white balls from the first bag or drawing exactly four white balls from bag 2.

We need to find the number of ways to order the drawing of white balls and then multiply it with the respective probability.

So, we get the required probability as \[\left( {{P}_{1}}{{P}_{w1}}_{\_4}{{P}_{b1}}_{\_1} \right)+\left( {{P}_{2}}{{P}_{w2}}_{\_4}{{P}_{b2}}_{\_1} \right)\].

$ \Rightarrow \left( {{P}_{1}}{{P}_{w1}}_{\_4}{{P}_{b1}}_{\_1} \right)+\left( {{P}_{2}}{{P}_{w2}}_{\_4}{{P}_{b2}}_{\_1} \right)=\left( \dfrac{1}{2}\times {}^{5}{{C}_{4}}\times {{\left( \dfrac{1}{2} \right)}^{4}}\times \left( \dfrac{1}{2} \right) \right)+\left( \dfrac{1}{2}\times {}^{5}{{C}_{4}}\times {{\left( \dfrac{3}{8} \right)}^{4}}\times \left( \dfrac{5}{8} \right) \right) $ .

$ \Rightarrow \left( {{P}_{1}}{{P}_{w1}}_{\_4}{{P}_{b1}}_{\_1} \right)+\left( {{P}_{2}}{{P}_{w2}}_{\_4}{{P}_{b2}}_{\_1} \right)=\left( \dfrac{5}{{{2}^{2}}}\times {{\left( \dfrac{1}{2} \right)}^{4}} \right)+\left( \dfrac{5}{16}\times {{\left( \dfrac{3}{8} \right)}^{4}} \right) $ .

$ \Rightarrow \left( {{P}_{1}}{{P}_{w1}}_{\_4}{{P}_{b1}}_{\_1} \right)+\left( {{P}_{2}}{{P}_{w2}}_{\_4}{{P}_{b2}}_{\_1} \right)=\left( \dfrac{5}{{{2}^{6}}} \right)+\left( \dfrac{5\times {{3}^{4}}}{{{2}^{16}}} \right) $ .

\[\Rightarrow \left( {{P}_{1}}{{P}_{w1}}_{\_4}{{P}_{b1}}_{\_1} \right)+\left( {{P}_{2}}{{P}_{w2}}_{\_4}{{P}_{b2}}_{\_1} \right)=\left( \dfrac{5}{{{2}^{6}}} \right)\left( 1+\dfrac{{{3}^{4}}}{{{2}^{10}}} \right)\].

$ \therefore $ We have found the required probability as $ \left( \dfrac{5}{{{2}^{6}}} \right)\left( 1+\dfrac{{{3}^{4}}}{{{2}^{10}}} \right) $ .

**Note:**

Here we have assumed that the bag is not replaced once choosing it as this will change the solution differently. We should not make calculation mistakes while solving this problem. We have assumed that drawing the balls for each item is an independent event. Similarly, we can expect problems to find the probability to draw 5 blacks in the given order.

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