JEE Advanced 2013 · Paper 1 · Mathematics Q58

← JEE Advanced 2013 · Paper 1

MathematicsInteger+4 / -1Question 58 of 60

Of the three independent events $E_1$ , $E_2$ and $E_3$ , the probability that only $E_1$ occurs is $\alpha$ , only $E_2$ occurs is $\beta$ and only $E_3$ occurs is $\gamma$ . Let the probability $p$ that none of events $E_1$ , $E_2$ or $E_3$ occurs satisfy the equations $(\alpha-2\beta)p=\alpha\beta$ and $(\beta-3\gamma)p=2\beta\gamma$ . All the given probabilities are assumed to lie in the interval $(0,1)$ .

Then $\dfrac{\text{Probability of occurrence of }E_1}{\text{Probability of occurrence of }E_3}=$

Your answer

Answer and worked solution

Official answer: 6

Variables. With $P(E_i)=x,y,z$ : $\alpha=x(1-y)(1-z)$ , $\beta=(1-x)y(1-z)$ , $\gamma=(1-x)(1-y)z$ , $p=(1-x)(1-y)(1-z)$ . So $\frac\alpha p=\frac x{1-x}$ , etc.

First equation divided by $\alpha\beta$ : $\frac p\beta-\frac{2p}\alpha=1$ , i.e. $\left(\frac1y-1\right)-2\left(\frac1x-1\right)=1$ , so $x=2y$ .

Second equation divided by $\beta\gamma$ : $\frac p\gamma-\frac{3p}\beta=2$ , i.e. $\left(\frac1z-1\right)-3\left(\frac1y-1\right)=2$ , so $y=3z$ .

Ratio $\dfrac xz=6$ .

Answer: 6

Topics:Independent events

← Previous question Next question →

JEE Advanced 2013 · Paper 1, Mathematics question 58