Question

The probability that it will rain tomorrow is $0.25$ . If it rains tomorrow, the probability that Amanda plays tennis is $0.1$ . If it doesn't rain tomorrow, the probability that she plays tennis is $0.9$ . Let $A$ be the event 'rains tomorrow' and $B$ be the event 'plays tennis'. Complete the following tree diagram for the events $A and B$ .

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Accepted Answer

The probability that it will rain tomorrow is $0.25$ . If it rains tomorrow, the probability that Amanda plays tennis is $0.1$ . If it doesn't rain tomorrow, the probability that she plays tennis is $0.9$ .

Let $A$ be the event 'rains tomorrow' and $B$ be the event 'plays tennis'.

Therefore, the $P (A) = 0.25$ and $P (A^{'}) = 1 - P (A) = 1 - 0.25 = 0.75$

If it rains tomorrow, the probability that Amanda plays tennis is $P (B ∣ A) = 0.1$

Therefore, $P (B ∣ A) = \frac{P (A \cap B)}{P (A)}$

$\Rightarrow P (A \cap B) = P (B ∣ A) \times P (A)$

$\Rightarrow P (A \cap B) = 0.1 \times 0.25$

$\Rightarrow P (A \cap B) = 0.025$

Now, $P (B^{'} ∣ A) = \frac{P (A \cap B^{'})}{P (A)} = 0.9$ [If it rains tomorrow, the probability that Amanda doesn't play tennis]

$\Rightarrow P (A \cap B^{'}) = 0.9 \times 0.25 = 0.225$

Also, $P (B ∣ A^{'}) = \frac{P (A^{'} \cap B)}{P (A^{'})} = 0.9$ [If it doesn't rain tomorrow, the probability that Amanda plays tennis]

$\Rightarrow P (A^{'} \cap B) = P (A^{'}) \times P (B ∣ A^{'})$

$\Rightarrow P (A^{'} \cap B) = 0.75 \times 0.9 = 0.675$

Similarly, $P (B^{'} ∣ A^{'}) = \frac{P (A^{'} \cap B^{'})}{P (A^{'})} = 0.1$ [If it doesn't rain tomorrow, the probability that Amanda doesn't play tennis]

$\Rightarrow P (A^{'} \cap B^{'}) = 0.1 \times 0.75 = 0.075$

Hence, the required tree diagram is

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