3. Discrete Random Variable _ Binomial Distribution Solutions 2023 (RVHS)

River Valley High School, Mathematics Department 2023 Discrete Random Variable & Binomial Distribution T1 Discrete Random Variable & Binomial Distribution 1(i) 11(1) (2) (3) (4) 140 40 40 40 kk + + + = 3 7 40k + = 11 [Shown]k= (ii) ( ) 33 P( ) P( ) 3 40 4 3 2 3 0P 2 3 4P( ) P( 3| 2 2 2, )3, 4 29140 40 40 40 XXXX XX X kk =   == == = = = = + + − (iii) 11 22 3 4E( ) 1 2 3 440 40 40 40X =  +  +  +  2= 2(i) 3 blue counters, 1 red counter and y yellow counters. S = No. of blue counters + 2(No. of red counters) P(S = 3) = P(BBB) + P(RBY in any order) 3 2 1 1 3 3!4 3 2 4 3 2 y y y y y y y=   +   + + + + + + 18 6 ( 4)( 3)( 2) y y y y += + + + [Shown] 3 13 3 1 1 1 44 33 or y yy C C C C CC++ + (ii) P( 73) 20S == 18 6 7 ( 4)( 3)( 2) 20 y y y y += + + + 32Expand and simplify: 7 63 178 48 0y y y+ − + = Since y is a positive integer, y = 2 from GC. Now we have 3 blue, 1 red counter and 2 yellow counters.

River Valley High School, Mathematics Department 2023 Discrete Random Variable & Binomial Distribution T2 Possible values of S are 1(YYB in any order), 2(BBY or RYY in any order), 3 (BBB or RBY in any order) and 4 (BBR in any order) P(S = 1) = P(YYB in any order) 2 1 3 3! 3 6 5 4 2! 20=    = P(S = 2) = P(BBY in any order) + P(RYY in any order) 3 2 2 3! 1 2 1 3! 7 6 5 4 2! 6 5 4 2! 20=    +    = P(S = 4) = P(BBR in any order) 3 2 1 3! 3 6 5 4 2! 20=    = The probability of S is s 1 2 3 4 P( )Ss= 3 20 7 20 7 20 3 20 3(i) x 0 1 2 3 4 P(X = x) 2k k 0 k 2k 2 2 1k k k k+ + + = 61k = 1 6k = 2 1 1 2E( ) 0 1 2 0 3 4 26 6 6 6X =  +  +  +  +  = Or By symmetry, E(X) = 2 (ii) E( 2 )X − all | 2 | P( ) 2 1 1 20 2 1 2 0 3 2 4 26 6 6 6 x x X x= − = = −  + −  + + −  + −   5 3= (iii)   22Var( ) E( ) E( )X X X=− 2 2 2 21 1 20 1 0 3 4 26 6 6       = + + + + −            1 9 32 46 6 6 = + + − 3=

River Valley High School, Mathematics Department 2023 Discrete Random Variable & Binomial Distribution T3 4 Question (i) ( )P different colour = ( ) 11P , 2 2 1 2 2 1 n n nBW n n n ++=   =++ . Question (ii) ( )P same colour ( ) ( )P , P ,B B W W=+ 1 1 1 1 2 1 2 2 1 2 2(2 1) 2(2 1) 2 1 n n n n n n n n n n n n n n − + − +=  +  = + =+ + + + + Question (iii) ( )P1X = ( ) 24 1 1 1 4! 3 2(2)2 1 2 2 1 2 3! 4 2 1 n n n n n n ++     = + =     + + +      Alternatively, ( )P1X = ( ) ( ) ( )P , , , P , , , , , P , , , , ,B W H T B B H T T T W W H T T T= + +