The probbltys of 3 mutually exclusive events are respectively (1-2p)/2,(1-p)/2,(1+3p)/3

Then:

a)1/3< p<1/2

b)0<p<1/2

c)1/3<p<2/3

d)0<p<2/3

Plz explain...

Question

The probbltys of 3 mutually exclusive events are respectively (1-2p)/2,(1-p)/2,(1+3p)/3 Then: a)1/3 < p < 1/2 b)0 < p < 1/2 c)1/3 < p < 2/3 d)0 < p < 2/3 Plz explain...

Chetan Mandayam Nayakar · Answer

1-2p&ge;0,p&le;1/2, similarly p&le;1 and p&ge;-(1/3), 1-2p&le;2,p>= -(1/2), similarly,p>= -1 and p&le;2/3 so p&le;(1/2) and p&ge;-(1/3), thus -(1/3)&le;p&le;(1/2)

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The probbltys of 3 mutually exclusive events are respectively (1-2p)/2,(1-p)/2,(1+3p)/3 Then: a)1/3 p 1/2 b)0 p 1/2 c)1/3 p 2/3 d)0 p 2/3 Plz explain...

The probbltys of 3 mutually exclusive events are respectively (1-2p)/2,(1-p)/2,(1+3p)/3

Then:

a)1/3< p<1/2

b)0<p<1/2

c)1/3<p<2/3

d)0<p<2/3

Plz explain...

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The probbltys of 3 mutually exclusive events are respectively (1-2p)/2,(1-p)/2,(1+3p)/3 Then: a)1/3 p 1/2 b)0 p 1/2 c)1/3 p 2/3 d)0 p 2/3 Plz explain...

The probbltys of 3 mutually exclusive events are respectively (1-2p)/2,(1-p)/2,(1+3p)/3 Then: a)1/3< p<1/2 b)0<p<1/2 c)1/3<p<2/3 d)0<p<2/3 Plz explain...

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The probbltys of 3 mutually exclusive events are respectively (1-2p)/2,(1-p)/2,(1+3p)/3

Then:

a)1/3< p<1/2

b)0<p<1/2

c)1/3<p<2/3

d)0<p<2/3

Plz explain...