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2 years ago

When a coin and die are tossed together find the probability of getting:

Mathematics

1 answer:

postnew [5]2 years ago

7 0

Answer:

a) 1/4

b) 1/6

c) 1/12

Step-by-step explanation:

Let S be the sample space.

S={H1,H2,H3,H4,H5,H6,T1,T2,T3,T4,T5,T6}

n(s) =12

<em><u>Events </u></em>

A: coin with head and die with prime number .

B:coin with head and die with composite number.

C:coin with tail and die with even prime number.

a) A={H2,H3,H5}

n(A) = 3

P(A) =n(A)/n(S)

=3/12

= 1/4

b) B={H4,H6}

n(B)= 2

P(B) = n(B)/n(S)

= 2/12

= 1/6

c) C ={T2}

n(C) = 1

P(C) = n(C)/n(S)

= 1/12

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In this case, the normal vector is

$\dfrac{\partial\vec r}{\partial x}\times\dfrac{\partial\vec r}{\partial y}=-\dfrac{\partial(8-x-2y)}{\partial x}\,\vec\imath-\dfrac{\partial(8-x-2y)}{\partial y}\,\vec\jmath+\vec k=\vec\imath+2\,\vec\jmath+\vec k$

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$\displaystyle\iint_Se^z\,\mathrm d\Sigma=\sqrt6\iint_Te^{8-x-2y}\,\mathrm dy\,\mathrm dx$

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