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Let S, be event be 'he must stop ot sighal 1" s_{2} be event be "he must stop at sighall"Given\begin{array}{l}P\left(s_{1}\right)=0.4 \\P\left(S_{2}\right)=0.5\end{array}probability that he must stopatleost on of Siagnal P\left(S_{1} \cup S_{2}\right)=0.79)\text { At both sighals } \Rightarrow P\left(S_{1} \cap S_{2}\right)\Rightarrow wetsinow that P\left(S_{1} \cup S_{2}\right)=P\left(s_{1}\right)+\beta\left(S_{2}\right)-P\left(S_{1} \cap S_{2}\right)\begin{array}{l}0.7=0.4+0.5-P\left(S_{1} \cap S_{2}\right) \\P\left(S_{1} \cap S_{2}\right)=0.9-0.7=0.2\end{array}\Rightarrow P\left(S_{1} \cap S_{2}\right)=0.9-0.7=0.2\because P\left(s_{1} \cap s_{2}\right)=0.2b) only at signal 1 but not sighal 2\Rightarrow P\left(S_{1} \cap S_{2}^{1}\right)S_{2}^{\prime}.implies complementaly event of S_{2}A C C to theorm of total probability\begin{array}{l}P\left(S_{1}\right)=P\left(S_{1} \cap S_{2}\right)+P\left(S_{1} \cap S_{2}^{1}\right) \\0.4=0.2+P\left(S_{1} \cap S_{2}^{1}\right) \\\Rightarrow P\left(S_{1} \cap S_{2}^{1}\right)=0.2\end{array}c) Probability of stopping at exactly one of sighal. is sumot probability of stopping ouly at ist siagual and probability of stopping only of 2^{\text {nd }} sighal.\Rightarrow P\left(S_{1} \cap S_{2}^{1}\right)+P\left(S_{1} \cap S_{2}\right)we know value of \beta\left(S, A S_{2}^{\prime}\right) we will find P\left(S_{1}^{\prime} \cap S_{2}\right) as above.\begin{array}{l}P\left(S_{2}\right)=P\left(S_{1} \cap S_{2}\right)+P\left(S_{1}^{\prime} \cap S_{2}\right) \\P\left(S_{1}^{\prime} \cap S_{2}\right)=0.5-0.2=0.3\end{array}\therefore Required Probability\begin{aligned}P\left(s_{1}^{\prime} \cap s_{2}\right)+P\left(s_{1} \cap s_{2}^{\prime}\right) & =0.2+0.3 \\& =0.5 .\end{aligned} ...