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Probability — MATHS STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceMATHSProbability2 Marks
Question
There are three events A, B, C one of which must and only one can happen, the odds are 8 to 3 against A, 5 to 2 against B, find the odds against C.
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Answer

Given, $\text{P}(\overline{\text{A}}):\text{P}(\text{B})=8:3$ $\Rightarrow\frac{1-\text{P}(\text{A})}{\text{P}(\text{A})}=\frac{8}{3}$ $\Rightarrow\text{P}(\text{A})=\frac{3}{11}$ $\text{P}(\overline{\text{B}}):\text{P}(\text{B})=5:2$ $\Rightarrow\frac{1-\text{P}(\text{B})}{\text{P}(\text{B})}=\frac{5}{2}$ $\Rightarrow\frac{1}{\text{P}(\text{B})}=\frac{5}{2}+1=\frac{7}{2}$ $\Rightarrow\text{P}(\text{B})=\frac{2}{7}$ $\because$ A, B and C are mutually exhaustive $\therefore\text{A}\cup\text{B}\cup\text{C}=\text{S}$ $\Rightarrow\text{P}(\text{A}\cup\text{B}\cup\text{C})=\text{P}(\text{S})$ $\Rightarrow\text{P}(\text{A})+\text{P}(\text{B})+\text{P}(\text{C})=1$ $\text{P}(\text{C})=1-\big\{\text{P}(\text{A})+\text{P}(\text{B})\big\}$ $=1-\Big(\frac{3}{11}+\frac{2}{7}\Big)$ $=1-\frac{43}{77}$ $=\frac{34}{77}$ $\Rightarrow\text{P}(\overline{\text{C}})=1-\text{P}(\text{C})$ $=1-\frac{34}{77}$ $=\frac{43}{77}$ $\therefore$ Odds against C is $\text{P}(\overline{\text{C}}):\text{P}(\text{C})=\frac{43}{77}:\frac{34}{77}$ $=43:34$

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