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Probability — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsProbability4 Marks
Question
Prove that:
  1. $\text{P}(\text{A})=\text{P}(\text{A}\cap\text{B})+\text{P}(\text{A}\cap\bar{\text{B}})$
  2. $\text{P}(\text{A}\cup\text{B})=\text{P}(\text{A}\cap\text{B})+\text{P}(\text{A}\cap\bar{\text{B}})+\text{P}(\bar{\text{A}}\cap\text{B})$
✓

Answer

  1. $\because\text{P}(\text{A})=\text{P}(\text{A}\cap\text{B})+\text{P}(\text{A}\cap\bar{\text{B}})$

$\therefore\text{R.H.S.}=\text{P}(\text{A}\cap\text{B})+\text{P}(\text{A}\cap\bar{\text{B}})$

$=\text{P}(\text{A})\cdot\text{P}(\text{B})+\text{P}(\text{A})\cdot\text{P}\bar{(\text{B})}$

$=\text{P}(\text{A})\big[\text{P}(\text{B})+\text{P}\bar{(\text{B})}\big]$

$=\text{P}(\text{A})\big[\text{P}(\text{B})+1-\text{P}(\text{B})\big]$ $\big[\because\text{P}\bar{(\text{B})}=1-\text{P}(\text{B})\big]$

$= \text{P(A) = L. H. S} $ Hence proved.

  1. $\because\text{P}(\text{A}\cup\text{B})=\text{P}(\text{A}\cap\text{B})+\text{P}(\text{A}\cap\bar{\text{B}})+\text{P}(\bar{\text{A}}\cap\text{B})$

$\therefore\text{R.H.S.}=\text{P}(\text{A})\cdot\text{P}(\text{B})+\text{P}(\text{A})\cdot\text{P}\bar{(\text{B})}+\text{P}\bar{(\text{A})}\cdot\text{P}(\text{B})$

$=\text{P}(\text{A})\cdot\text{P}(\text{B})+\text{P}(\text{A})\cdot\big[1-\text{P}(\text{B})\big]+\big[1-\text{P}(\text{A})\big]\text{P}(\text{B})$

$=\text{P}(\text{A})\cdot\text{P}(\text{B})+\text{P}(\text{A})-\text{P}(\text{A})\cdot\text{P}(\text{B})+\text{P}(\text{B})-\text{P}(\text{A})\cdot\text{P}(\text{B})$

$=\text{P}(\text{A})+\text{P}(\text{B})-\text{P}(\text{A})\cdot\text{P}(\text{B})$

$=\text{P}(\text{A})+\text{P}(\text{B})-\text{P}(\text{A}\cap\text{B})$

$=\text{P}(\text{A}\cup\text{B})=\text{L.H.S.}$ Hence proved.

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