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Question

(A) Let $p_1$ and $p_2$ be the probabilities of the first and second events respectively. According to the given conditions: $p_1 = p_2^2$ The odds ag

Vedclass · Accepted Answer

$\frac{1}{9}, \frac{1}{3}$

Vedclass · Answer

(A) Let $p_1$ and $p_2$ be the probabilities of the first and second events respectively. According to the given conditions: $p_1 = p_2^2$ The odds against an event with probability $p$ are given by $\frac{1-p}{p}$. Therefore,$\frac{1-p_1}{p_1} = \left(\frac{1-p_2}{p_2}ight)^3$. Substituting $p_1 = p_2^2$: $\frac{1-p_2^2}{p_2^2} = \frac{(1-p_2)^3}{p_2^3}$ $\frac{(1-p_2)(1+p_2)}{p_2^2} = \frac{(1-p_2)^3}{p_2^3}$ Assuming $p_2 
eq 1$ and $p_2 
eq 0$,we divide by $\frac{1-p_2}{p_2^2}$: $1+p_2 = \frac{1-p_2}{p_2}$ $p_2 + p_2^2 = 1 - p_2$ $p_2^2 + 2p_2 - 1 = 0$ Solving for $p_2$ using the quadratic formula: $p_2 = \frac{-2 \pm \sqrt{4 - 4(1)(-1)}}{2} = \frac{-2 \pm \sqrt{8}}{2} = -1 \pm \sqrt{2}$. Since $0 \le p_2 \le 1$,we take $p_2 = \sqrt{2} - 1$. This does not match the provided options. Re-evaluating the problem statement: if the odds against the first are the cube of the odds against the second,and $p_1 = 1/9, p_2 = 1/3$: Odds against $p_1 = \frac{1-1/9}{1/9} = 8$. Odds against $p_2 = \frac{1-1/3}{1/3} = 2$. Since $8 = 2^3$,the condition holds. Thus,the correct option is $A$.

The probability of an event happening is the square of the probability of a second event,but the odds against the first are the cube of the odds against the second. The probabilities of the events are:

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