Question

Given two independent events A and B such that P(A) = 0.3, P(B) = 0.6. Find
(i) P(A and B)
(ii) P(A and not B)
(iii) P(A or B)
(iv) P(neither A nor B)

Answer 1

Answer:

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Step 3: Refocus. ...

Step 4: Revalu

Step-by-step explanation:

Answer 2

Given two independent events A and B such that P(A) = 0.3, P(B) = 0.6. the value of

(i) P(A and B)  = 0.18

(ii) P(A and not B) =0.12

(iii) P(A or B) =0.72

(iv) P(neither A nor B)=0.28

Step-by-step explanation:

Given $\textrm{P}(\textrm{A})=0.3, \textrm{P}(\textrm{B})=0.6$

A and B are Independent events

i.e $\textrm{P}(\textrm{A}\textrm{and}\textrm{B})=\textrm{P}(\textrm{A}). \textrm{P}(\textrm{B})$

(I) $\textrm{P}(\textrm{A}\textrm{and}\textrm{B})=\textrm{P}(\textrm{A}). \textrm{P}(\textrm{B})=0.3 \times0.6=0.18$

(ii) $\textrm{P}(\textrm{A}\textrm{and}\textrm {notB})=\textrm{P}(\textrm{A}). \textrm{P}(\textrm{notB})=\textrm{P}(\textrm{A}). (1-\textrm{P}(\textrm{B}))$

  $=0.3 (1-0.6)$

  $=0.3 \times0.4$

 $=0.12$

(iii) $\textrm{P}(\textrm{A}\textrm {Or}\textrm{B})$

    $=\textrm{P}(\textrm{A}\bigcup \textrm{B})$

$=\textrm{P}(\textrm{A})+\textrm{P}(\textrm{B})-\textrm{P}(\textrm{A}\bigcap \textrm{B})$

  $=0.3+0.6-0.18$

  $=0.72$

(iv) $\textrm{P}(\textrm{neither }\textrm{A}\textrm {nor}\textrm {B})=\textrm{P}\overline{(\textrm{A}\bigcup\textrm{B)} }=1-\textrm{P}(\textrm{A}\bigcup \textrm{B})$

 $=1-0.72$

$=0.28$