The sample space consist of 6 * 4 = 24 sample points as:
{1, 1}; {1, 2}; {1, 3}; {1, 4} ;
{2, 1}; {2, 2}; {2, 3}; {2, 4} ;
{3, 1}; {3, 2}; {3, 3}; {3, 4} ;
{4, 1}; {4, 2}; {4, 3}; {4, 4} ;
{5, 1}; {5, 2}; {5, 3}; {5, 4} ;
{6, 1}; {6, 2}; {6, 3}; {6, 4}
a) A = { sum of 5 is rolled}
therefore A consist of {1, 4}; { 2,3}, {3, 2}; {4,1}
Therefore P(A) = 4/24 = 1/6 = 0.166667
b) B = {sum of 5 or sum of 9 is rolled}.
So B = { {1, 4}; { 2,3}, {3, 2}; {4,1}; {5, 4}; {6, 3} }
Therefore P( B ) = 6/24 = 1/4 = 0.25
c) P( A and B ) = P( Common sample points in A and B ) = 4/24 = 1/6 = 0.166667
pun LU I LLULU IUF this question. Enter electrons as e Use smallest possible integer coefficients. If a box is not needed, leave it blank. For the following electron-transfer reaction: 2Cr(s) + 31,(s) — 2Cr3+(aq) + 61'(aq) The oxidation half-reaction is: The reduction half-reaction is: Submit Answer Retry Entire Group 1 more group attempt remaining Provious Ne
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PROBLEM 5 Starting with the integral equation of motion, Ot derive the differential form of the equation. Hint: To do this, look at how we derived the differential form of the mass continuity equation. There are parallels, although thisis more complicated. Note that youil ave to apply the gradien identt. fHp di -