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Probability — MATHS STD 11 Science — Question

Rajasthan BoardEnglish MediumSTD 11 ScienceMATHSProbability5 Marks
Question
A sample space consists of 9 elementary outcomes e1, e2, ...., e9 whose probabilities are
P(e1) = P(e2 ) = 0.08, P(e3 ) = P(e4) = P(e5) = 0.1
P(e6) = P(e7) = 0.2, P(e8) = P(e9) = 0.07
Suppose A = {e1, e5, e8}, B = {e2, e5, e8, e9}
  1. Calculate P (A), P (B), and $\text{P}(\text{A}\cap\text{B})$
  2. Using the addition law of probability, calculate $\text{P}(\text{A}\cup\text{B})$
  3. List the composition of the event $\text{A}\cup\text{B},$ and calculate $\text{P}(\text{A}\cup\text{B})$ by adding the probabilities of the elementary outcomes.
  4. Calculate $\text{P}(\bar{\text{B}})$ from P(B), also calculate $\text{P}(\bar{\text{B}})$ directly from the elementary outcomes of $\bar{\text{B}}$
✓

Answer

  1. $\text{P(A)}=\text{P}(\text{E}_1)+\text{P}(\text{E}_5)+\text{P}(\text{E}_8)$

$\text{P(A)}=0.08+0.1+0.07=0.25$

  1. $\text{P(B)}=\text{P}(\text{E}_2)+\text{P}(\text{E}_5)+\text{P}(\text{E}_8)+\text{P}(\text{E}_9)$

$\text{P(B)}=0.08+0.1+0.07+0.07=0.32$

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

Now, $\text{A}\cap\text{B}=\big\{\text{E}_5,\text{E}_8\big\}$

$\therefore\ \text{P}(\text{A}\cap\text{B})=\text{P}(\text{E}_5)+\text{P}(\text{E}_8)$

$=0.1+0.07=0.17$

$\therefore\ \text{P}(\text{A}\cup\text{B})=0.25+0.32-0.17=0.40$

  1. $\text{A}\cup\text{B}=\big\{\text{E}_1,\text{E}_2,\text{E}_5,\text{E}_8,\text{E}_9\big\}$

$\text{P}(\text{A}\cup\text{B})=\text{P}(\text{E}_1)+\text{P}(\text{E}_2)+\text{P}(\text{E}_5)+\text{P}(\text{E}_8)+\text{P}(\text{E}_9)$

$=0.08+0.08+0.1+0.07+0.07=0.40$

  1. $\because\ \text{P}\bar{(\text{B})}=1-\text{P(B)}$

$=1-0.32=0.68$

and $\bar{\text{B}}=\big\{\text{E}_1,\text{E}_3,\text{E}_4,\text{E}_6,\text{E}_7\big\}$

$\therefore\ \text{P}(\bar{\text{B}})=\text{P}(\text{E}_1)+\text{P}(\text{E}_3)+\text{P}(\text{E}_4)+\text{P}(\text{E}_6)+\text{P}(\text{E}_7)$

$=0.08+0.1+0.1+0.2+0.2=0.68$

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