Let B_i (i = 1, 2, 3) be three independent events in a sample space. The probability that only B₁ occurs is $γ .$ a, only B₂ occurs is b and B₃ occurs is Let p be the probability that none of the events B₁ occurs and these 4 probabilities satisfy the equations (a - 2b) = ab and (b - 3 $γ)$ p = $2 β γ$ (All the probabilities are assumed to lie in the integral (0, 1)). The $\frac{P (B_{1})}{P (B_{3})}$ is equal to_________.

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alok kumar singh | Contributor-Level 10

a month ago

Let P (B₁) = a P (B₂) = b P (B₃) = c

Given a (1 – b) (1 – c) = a . (i)

b (1 – a) (1 – c) = b . (ii)

c (1 – b) (1 – a) = $γ$ . (iii)

(1 – a) (1 – b) (1 – c) = p . (iv)

$(α - 2 β) p = α β$

->a – ab – 2b + 2ab = ab Þ a = 2b . (v)

Again $(β - 3 γ) p = 2 β γ$

->b – bc – 3c + 3bc = 2bc Þ b = 3c . (vi)

$\Rightarrow \frac{P (B_{1})}{P (B_{3})} = \frac{a}{c} = \frac{2 b}{b / 3} = 6$

...more

Let P (B₁) = a P (B₂) = b P (B₃) = c

Given a (1 – b) (1 – c) = a . (i)

b (1 – a) (1 – c) = b . (ii)

c (1 – b) (1 – a) = $γ$ . (iii)

(1 – a) (1 – b) (1 – c) = p . (iv)

$(α - 2 β) p = α β$

->a – ab – 2b + 2ab = ab Þ a = 2b . (v)

Again $(β - 3 γ) p = 2 β γ$

->b – bc – 3c + 3bc = 2bc Þ b = 3c . (vi)

$\Rightarrow \frac{P (B_{1})}{P (B_{3})} = \frac{a}{c} = \frac{2 b}{b / 3} = 6$

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Let P (B1) = a      P (B2) = b            P (B3) = cGiven a (1 – b) (1 – c) = a . (i)b (1 – a) (1 – c) = b              . (ii)c (1 – b) (1 – a) = <math> <mrow> <mi>γ</mi> </mrow> </math> <!- [if gte mso 9]><xml> <o:OLEObject Type="Embed" ProgID="Equation.DSMT4" ShapeID="_x0000_i1025" DrawAspect="Content" ObjectID="_1816675496"> </o:OLEObject></xml><! [endif]->            . (iii) (1 – a) (1 – b) (1 – c) = p     . (iv) <math> <mrow> <mrow> <mo> (</mo> <mrow> <mi>α</mi> <mo>−</mo> <mn>2</mn> <mi>β</mi> </mrow> <mo>)</mo> </mrow> <mi>p</mi> <mo>=</mo> <mi>α</mi> <mi>β</mi> </mrow> </math>  <!- [if gte mso 9]><xml> <o:OLEObject Type="Embed" ProgID="Equation.DSMT4" ShapeID="_x0000_i1026" DrawAspect="Content" ObjectID="_1816675497"> </o:OLEObject></xml><! [endif]->->a – ab – 2b + 2ab = ab Þ a = 2b . (v)Again <math> <mrow> <mrow> <mo> (</mo> <mrow> <mi>β</mi> <mo>−</mo> <mn>3</mn> <mi>γ</mi> </mrow> <mo>)</mo> </mrow> <mi>p</mi> <mo>=</mo> <mn>2</mn> <mi>β</mi> <mi>γ</mi> </mrow> </math>  <!- [if gte mso 9]><xml> <o:OLEObject Type="Embed" ProgID="Equation.DSMT4" ShapeID="_x0000_i1027" DrawAspect="Content" ObjectID="_1816675498"> </o:OLEObject></xml><! [endif]->->b – bc – 3c + 3bc = 2bc Þ b = 3c   . (vi) <math> <mrow> <mo>⇒</mo> <mfrac> <mrow> <mi>P</mi> <mrow> <mo> (</mo> <mrow> <msub> <mrow> <mi>B</mi> </mrow> <mrow> <mn>1</mn> </mrow> </msub> </mrow> <mo>)</mo> </mrow> </mrow> <mrow> <mi>P</mi> <mrow> <mo> (</mo> <mrow> <msub> <mrow> <mi>B</mi> </mrow> <mrow> <mn>3</mn> </mrow> </msub> </mrow> <mo>)</mo> </mrow> </mrow> </mfrac> <mo>=</mo> <mfrac> <mrow> <mi>a</mi> </mrow> <mrow> <mi>c</mi> </mrow> </mfrac> <mo>=</mo> <mfrac> <mrow> <mn>2</mn> <mi>b</mi> </mrow> <mrow> <mi>b</mi> <mo>/</mo> <mn>3</mn> </mrow> </mfrac> <mo>=</mo> <mn>6</mn> </mrow> </math>     <!- [if gte mso 9]><xml> <o:OLEObject Type="Embed" ProgID="Equation.DSMT4" ShapeID="_x0000_i1028" DrawAspect="Content" ObjectID="_1816675499"> </o:OLEObject></xml><! [endif]->

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