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PSYCHO15rus [73]
3 years ago
5

The Venn diagram shows the results of two events resulting from rolling a number cube.

Mathematics
1 answer:
tankabanditka [31]3 years ago
6 0

Answer:

P(A|B)=\frac{2}{3}

P(A)*P(B)=\frac{1}{3}

P(A) =\frac{2}{3}

P(B) =\frac{1}{2}.

Step-by-step explanation:

We use the Venn diagram to calculate the desired probabilities.

Note that there are 6 possible results in the sample space

S = {1, 2, 3, 4, 5, 6}

Then note that in the region representing the intercept of A and B there are two possible values.

So

P (A\ and\ B) = \frac{2}{6} = \frac{1}{3}

In the region that represents event A there are 4 possible outcomes {4, 5, 1, 2}

So

P(A) = \frac{4}{6} = \frac{2}{3}

In the region that represents event B there are 3 possible outcomes {1, 2, 6}

So

P(B) = \frac{3}{6} = \frac{1}{2}.

Now

P(A | B)=\frac{P(A \ and\ B)}{P(B)}\\\\P(A | B)=\frac{\frac{1}{3}}{\frac{1}{2}}\\\\P(A|B)=\frac{2}{3}

P(A)*P(B)=\frac{2}{3}*\frac{1}{2}=\frac{1}{3}

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\frac{a}{p}\times (q-r)+\frac{b}{q}\times (r-p)+\frac{c}{r}\times (p-q)\neq 0  is proved for the sum of pth, qth and rth terms of an arithmetic progression are a, b,and c respectively.

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Given that the sum of pth, qth and rth terms of an arithmetic progression are a, b and c respectively.

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\frac{a}{p}\times (q-r)+\frac{b}{q}\times (r-p)+\frac{c}{r}\times (p-q)=0

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\frac{a}{p}\times (q-r)+\frac{b}{q}\times (r-p)+\frac{c}{r}\times (p-q)

=\frac{A+(p-1)D}{p}\times (q-r)+\frac{A+(q-1)D}{q}\times (r-p)+\frac{A+(r-1)D}{r}\times (p-q)

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=\frac{[Aq+pqD-Dq-Ar-prD+rD]\times qr+[Ar+rqD-Dr-Ap-pqD+pD]\times pr+[Ap+prD-Dp-Aq-qrD+qD]\times pq}{pqr}

=\frac{Arq^{2}+pq^{2} rD-Dq^{2} r-Aqr^{2}-pqr^{2} D+qr^{2} D+Apr^{2}+pr^{2} qD-pDr^{2} -Ap^{2}r-p^{2} rqD+p^{2} rD+Ap^{2} q+p^{2} qrD-Dp^{2} q-Aq^{2} p-q^{2} prD+q^{2}pD}{pqr}

=\frac{Arq^{2}-Dq^{2}r-Aqr^{2}+qr^{2}D+Apr^{2}-pDr^{2}-Ap^{2}r+p^{2}rD+Ap^{2}q-Dp^{2}q-Aq^{2}p+q^{2}pD}{pqr}

=\frac{Arq^{2}-Dq^{2}r-Aqr^{2}+qr^{2}D+Apr^{2} -pDr^{2}-Ap^{2}r+p^{2}rD+Ap^{2}q-Dp^{2}q-Aq^{2}p+q^{2}pD}{pqr}

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Therefore LHS\neq RHS

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Hence proved

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