Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • NCERT
  • Let E1 and E2 be two independent events such that p(E1) = p1 and P(E2) = p2. Describe in words of the events whose probabilities are (i)P1P2 (ii)(1-P1-2) P2 (iii)1-(1-P1) (1-P2) (iv)P1+P2-2P1P2

Let E_1 and E_2 be two independent events such that P(E_1) = p_1 and P(E_2) = p_2. Describe in words of the events whose probabilities are
\\(i) P_{1} P_{2} \\(ii) \left(1-P_{1}\right) P_{2} \\(iii) 1-\left(1-P_{1}\right)\left(1-P_{2}\right) \\(\mathrm{iv}) \mathrm{P}_{1}+\mathrm{P}_{2}-2 \mathrm{P}_{1} \mathrm{P}_{2}

Answers (1)

Solution

\\Given \mathrm{P}\left(\mathrm{E}_{1}\right)=\mathrm{p}_{1} and \mathrm{P}\left(\mathrm{E}_{2}\right)=\mathrm{p}_{2} \\\\(\mathrm{i}) \mathrm{p}_{1} \mathrm{p}_{2} =P\left(E_{1}\right) \cdot P\left(E_{2}\right)=P\left(E_{1} \cap E_{2}\right)

Hence, E_{1} and E_{2} occur simultaneously.

\\\\(ii)(1-p_1) \mathrm{p}_{2} =\mathrm{P}\left(\mathrm{E'}_{1}\right)\cdot \mathrm{P}\left(\mathrm{E}_{2}\right) =P\left(E_{1}^{\prime} \cap E_{2}\right)

Hence, \mathrm{E}_{1} does not occur but \mathrm{E}_{2} occurs

\\\\(iii) 1-\left(1-p_{1}\right) \left(1-p_{2}\right) =1-P\left(E'_{1}\right) P\left(E'_{2}\right) \\=1-P\left(E_{1}^{\prime} \cap E_{2}^{\prime}\right).

By De Morgan's laws
\begin{array}{l} (\mathrm{A} \cup \mathrm{B})^{\prime}=\mathrm{A}^{\prime} \cap \mathrm{B}^{\prime} \\ (\mathrm{A} \cap \mathrm{B})^{\prime}=\mathrm{A}^{\prime} \cup \mathrm{B}^{\prime} \\ \end{array}

Knowing that

\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{A})^{\prime}=1

\\=1-\left[1-P\left(E_{1} \cup E_{2}\right)\right] \\=P\left(E_{1} \cup E_{2}\right)
Hence, either E_{1} or E_{2} or both E_{1} and E_{2} occurs.

(iv)

\\\mathrm{p}_{1}+\mathrm{p}_{2}-2 \mathrm{p}_{1} \mathrm{p}_{2} \\\\=\mathrm{P}\left(\mathrm{E}_{1}\right)+\mathrm{P}\left(\mathrm{E}_{2}\right)-2 \mathrm{P}\left(\mathrm{E}_{1}\right) \mathrm{P}\left(\mathrm{E}_{2}\right) \\\\=\mathrm{P}\left(\mathrm{E}_{1}\right)+\mathrm{P}\left(\mathrm{E}_{2}\right)-2 \mathrm{P}\left(\mathrm{E}_{1} \cap \mathrm{E}_{2}\right) \\\\=\mathrm{P}\left(\mathrm{E}_{1} \cup \mathrm{E}_{2}\right)- \mathrm{P}\left(\mathrm{E}_{1} \cap \mathrm{E}_{2}\right)
Hence, either E_{1} or E_{2} occurs but not both.

Posted by

infoexpert22

View full answer

Similar Questions

Latest Question