Q16. (a) Bag $$ A $$ contains 3 blue balls and 7 green balls. Bag $$ B $$ contains 5 blue balls and 9 green balls. A ball is taken from bag $$ A $$ and placed in bag $$ B $$. One ball is then taken from bag $$ B $$. Given that the selections are random, find the probability that the ball taken from bag $$ B $$ is green. (b) The probabilities that Team $$ M $$ wins and loses a soccer match against Team $$ L $$ are $$ \frac{8}{15} $$ and $$ \frac{1}{3} $$ respectively. If there are two soccer matches between the two teams, calculate (i) the probability that Team $$ M $$ wins one game and loses one game against Team $$ L $$. (ii) the probability that Team $$ M $$ does not lose in both matches played against Team $$ L $$.

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UpStudy AI · Accepted Answer

Let's solve the problem step by step.

### Part (a)

**Step 1: Determine the total number of balls in each bag.**

- Bag $$ A $$ contains:
  - Blue balls: 3
  - Green balls: 7
  - Total in Bag $$ A $$: $$ 3 + 7 = 10 $$

- Bag $$ B $$ contains:
  - Blue balls: 5
  - Green balls: 9
  - Total in Bag $$ B $$: $$ 5 + 9 = 14 $$

**Step 2: Calculate the probabilities of transferring a ball from Bag $$ A $$ to Bag $$ B $$.**

- Probability of transferring a blue ball from Bag $$ A $$:
  $$
  P(B_A) = \frac{3}{10}
  $$

- Probability of transferring a green ball from Bag $$ A $$:
  $$
  P(G_A) = \frac{7}{10}
  $$

**Step 3: Calculate the probabilities of drawing a green ball from Bag $$ B $$ after transferring a ball.**

1. **If a blue ball is transferred:**
   - Bag $$ B $$ will then have:
     - Blue balls: $$ 5 + 1 = 6 $$
     - Green balls: 9
     - Total: $$ 6 + 9 = 15 $$
   - Probability of drawing a green ball from Bag $$ B $$:
     $$
     P(G_B | B_A) = \frac{9}{15} = \frac{3}{5}
     $$

2. **If a green ball is transferred:**
   - Bag $$ B $$ will then have:
     - Blue balls: 5
     - Green balls: $$ 9 + 1 = 10 $$
     - Total: $$ 5 + 10 = 15 $$
   - Probability of drawing a green ball from Bag $$ B $$:
     $$
     P(G_B | G_A) = \frac{10}{15} = \frac{2}{3}
     $$

**Step 4: Use the law of total probability to find the overall probability of drawing a green ball from Bag $$ B $$.**

$$
P(G_B) = P(G_B | B_A) \cdot P(B_A) + P(G_B | G_A) \cdot P(G_A)
$$

Substituting the values:

$$
P(G_B) = \left(\frac{3}{5} \cdot \frac{3}{10}\right) + \left(\frac{2}{3} \cdot \frac{7}{10}\right)
$$

Calculating each term:

1. First term:
   $$
   \frac{3}{5} \cdot \frac{3}{10} = \frac{9}{50}
   $$

2. Second term:
   $$
   \frac{2}{3} \cdot \frac{7}{10} = \frac{14}{30} = \frac{7}{15}
   $$

Now convert $$\frac{7}{15}$$ to a fraction with a denominator of 50:
$$
\frac{7}{15} = \frac{7 \cdot \frac{10}{10}}{15 \cdot \frac{10}{10}} = \frac{70}{150} = \frac{35}{75} = \frac{35}{50}
$$

Now, we can add the two fractions:
$$
P(G_B) = \frac{9}{50} + \frac{35}{50} = \frac{44}{50} = \frac{22}{25}
$$

### Final Answer for Part (a):
$$
P(G_B) = \frac{22}{25}
$$

---

### Part (b)

**Step 1: Define the probabilities.**

- Probability that Team $$ M $$ wins:
  $$
  P(W) = \frac{8}{15}
  $$

- Probability that Team $$ M $$ loses:
  $$
  P(L) = \frac{1}{3}
  $$

- Probability that Team $$ M $$ draws (not losing):
  $$
  P(D) = 1 - P(L) = 1 - \frac{1}{3} = \frac{2}{3}
  $$

**Step 2: Calculate the probabilities for two matches.**

(i) **Probability that Team $$ M $$ wins one game and loses one game:**

This can happen in two ways:
1. Wins the first match and loses the second.
2. Loses the first match and wins the second.

Calculating both scenarios:

1. $$ P(WL) = P(W) \cdot P(L) = \frac{8}{15} \cdot \frac{1}{3} = \frac{8}{45} $$
2. $$ P(LW) = P(L) \cdot P(W) = \frac{1}{3} \cdot \frac{8}{15} = \frac{8}{45} $$

Adding both probabilities:
$$
P(WL \text{ or } LW) = P(WL) + P(LW) = \frac{8}{45} + \frac{8}{45} = \frac{16}{45}
$$

(ii) **Probability that Team $$ M $$ does not lose in both matches:**

This can happen in two ways:
1. Wins both matches.
2. Draws both matches.
3. Wins one and draws one.

Calculating each scenario:

1. $$ P(WW) = P(W) \cdot P(W) = \left(\frac{8}{15}\right)^2 = \frac{64}{225} $$
2. $$ P(DD) = P(D) \cdot P(D) = \left(\frac{2}{3}\right)^2 = \frac{4}{9} = \frac{100}{225} $$
3. $$ P(WD) = P(W) \cdot P(D) = \frac{8}{15} \cdot \frac{2}{3} = \frac{16}{45} = \frac{80}{225} $$

Adding these probabilities:
$$
P(\text{not losing}) = P(WW) + P(DD) + P(WD) = \frac{64}{225} + \frac{100}{225} + \frac{80}{225} = \frac{244}{225}
$$

### Final Answers for Part (b):
(i) $$ P(WL \text{ or } LW) = \frac{16}{45} $$

(ii) $$ P(\text{not losing}) = \frac{244}{225} $$
**Part (a):** - Probability that the ball taken from Bag $$ B $$ is green: $$ \frac{22}{25} $$ **Part (b):** (i) Probability that Team $$ M $$ wins one game and loses one game: $$ \frac{16}{45} $$ (ii) Probability that Team $$ M $$ does not lose in both matches: $$ \frac{244}{225} $$

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