Given below are two statement based on the following If A and B are independent events such that P(A)=p, P(B)=2 p and P (exactly one of A, B)=5/9 Statement I: p=1/3 Statement II: p=5/12 In the light of the above statements, choose the correct answer form the question given below Option 1: Both Statement I and Statement II are true Option 2: Both Statement I and Statement II are false Option 3: Statement I is true but Statement II is false Option 4: Statement I is false but Statement II is true

Answer: Both Statement I and Statement II are true. Solution: For independent events: Given P(A) = p, P(B) = 2p P(exactly one) = 5/9 P(exactly one) = P(AB AB) = P(A)(1-P(B)) + P(B)(1-P(A)) Therefore: 5/9 = p(1-2p) + 2p(1-p) = p - 2p^2 + 2p - 2p^2 = 3p - 4p^2 4p^2 - 3p + 5/9 = 0 36p^2 - 27p + 5 = 0 For p = 1/3 (Statement I): 36(1/9) - 27(1/3) + 5 = 0 ✓ For p = 5/12 (Statement II): 36(25/144) - 27(5/12) + 5 = 0 ✓ Therefore, both statements are true.

Modern Math

Probability

Hard

Given below are two statement based on the following

If $A$ and $B$ are independent events such that $P(A)=p, P(B)=2 p$ and $P$ (exactly one of $A, B)=\frac{5}{9}$

Statement I: $\mathrm{p}=\frac{1}{3}$

Statement II: $\mathrm{p}=\frac{5}{12}$

In the light of the above statements, choose the correct answer form the question given below

Both Statement I and Statement II are true

Both Statement I and Statement II are false

Statement I is true but Statement II is false

Statement I is false but Statement II is true

✅ Correct Option: 1

For independent events:

Given $P(A) = p$ , $P(B) = 2p$

$P(\text{exactly one}) = \frac{5}{9}$

$P(\text{exactly one}) = P(A\bar{B} \cup \bar{A}B)$

$= P(A)(1-P(B)) + P(B)(1-P(A))$

Therefore: $\frac{5}{9} = p(1-2p) + 2p(1-p)$

$= p - 2p^2 + 2p - 2p^2$

$= 3p - 4p^2$

$\Rightarrow 4p^2 - 3p + \frac{5}{9} = 0$

$\Rightarrow 36p^2 - 27p + 5 = 0$

For $p = \frac{1}{3}$ (Statement I):

$\Rightarrow 36(\frac{1}{9}) - 27(\frac{1}{3}) + 5 = 0$ ✓

For $p = \frac{5}{12}$ (Statement II):

$\Rightarrow 36(\frac{25}{144}) - 27(\frac{5}{12}) + 5 = 0$ ✓

Therefore, both statements are true.

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