Question

Let $B_i(i=1,2,3)$ be three independent events in a sample space. The probability that only $B_1$occur is $\alpha$, only $B_2$ occurs is $\beta$ and only $B_2$ occurs is $\gamma$. Let $p$ be the probability that none of the events $B_i$ occurs and these $4$ probabilities satisfy the equations $\left(\alpha –2\beta \right)p=\alpha \beta$ and $\left(\beta –3\gamma \right)p=2\beta \gamma$(All the probabilities are assumed to lie in the interval $\left(0,1\right)$. Then $\frac{P\left(B_1\right)}{P\left(B_3\right)}$is equal to __________.

Answer 1

Step 1: Declare the variables and equations

Let $x,y,z$ be probability of $B_1,B_2,B_3$ respectively.

$$\begin{gathered} \Rightarrow x\left(1–y\right)\left(1–z\right)=\alpha \\ \Rightarrow y\left(1–x\right)\left(1–z\right)=\beta \\ \Rightarrow z\left(1–x\right)\left(1–y\right)=\gamma \\ \Rightarrow \left(1-x\right)\left(1–y\right)\left(1–z\right)=p \end{gathered}$$

Consider the given Equation as

$$\begin{gathered} \left(\alpha –2\beta \right)p=\alpha \beta ...\left(1\right) \\ \left(\beta –3\gamma \right)p=2\beta \gamma ...\left(2\right) \end{gathered}$$

Step 2: Determine the value of $x$and $y$ by using the Equation $\left(1\right)\mathrm{and}\left(2\right)$

$$\begin{gathered} \left(\alpha -2\beta \right)p=\alpha \beta \\ \Rightarrow [x\left(1–y\right)\left(1–z\right)-2y\left(1-x\right)\left(1–z\right)]\left[\left(1-x\right)\left(1-y\right)\left(1–z\right)\right]=\left[x\left(1–y\right)\left(1–z\right)\right]\left[y\left(1-x\right)\left(1–z\right)\right] \\ \Rightarrow \left(x+xy-2y\right)\left(1+xy-x-y\right)=xy\left(1–y\right)\left(1-x\right) \\ \Rightarrow x+xy-2y=xy \\ \Rightarrow x=2y...\left(3\right) \end{gathered}$$

Similarly,

$$\begin{gathered} \left(\beta +3\gamma \right)p=2\beta \gamma \\ \Rightarrow \left(y\left(1–x\right)\left(1–z\right)-3z\left(1-x\right)\left(1–y\right)\left(\left(1–x\right)\left(1–y\right)\left(1–z\right)\right)\right)=yz{\left(1–x\right)}^2\left(1–y\right)\left(1–z\right) \\ \Rightarrow y+yz-3z=yz \\ \Rightarrow y=3z...\left(4\right) \end{gathered}$$

Substitute the value of $y$ in Equation $\left(3\right)$

$$\begin{gathered} \Rightarrow x=2y \\ \Rightarrow x=2\left(3z\right) \\ \Rightarrow x=6z \end{gathered}$$

Step 3: Write the value of $x,y,z$ in the form of probability.

$$\begin{gathered} x=6z \\ \Rightarrow B_1=6B_3 \\ \Rightarrow P\left(B_1\right)=6P\left(B_3\right) \\ \Rightarrow \frac{P\left(B_1\right)}{P\left(B_3\right)}=6 \end{gathered}$$

Hence, $\frac{P\left(B_1\right)}{P\left(B_3\right)}=6$.