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= + + ( ) 2 4 4 1 1 1 1 4! 1 1 4! 3 2(2)2 1 2 2(2 1) 2 3! 2(2 1) 2 3! 4 2 1 n n n n n n n n    + − + +      = + + =          + + + +           Question (iv) If there are 3 blue cards 3n= ( ) 14P, 2 1 7 nBW n +== + ; ( ) ( )1 1 1 2P , ;P , 2(2 1) 7 2(2 1) 7 nnB B W W nn −+= = = =++ ( ) 3 2 11P1 4(2 1) 28 nX n += = = + ( ) ( ) ( ) 44 P 4 P , , , , , P , , , , , 1 1 2 1 3 7 2 7 2 112 X B B H H H H W W H H H H= = +    = + =       ( ) ( ) ( ) 44 P 3 P , , , , , P , , , , , 1 1 4! 2 1 4! 3 7 2 3! 7 2 3! 28 X B B H H H T W W H H H T= = +        = + =               ( ) ( ) ( ) ( ) 244 P 0 P , , , P , , , , , P , , , , , 4 1 1 1 2 1 19 7 2 7 2 7 2 112 X B W T T B B T T T T W W T T T T= = + +      = + + =          

River Valley High School, Mathematics Department 2023 Discrete Random Variable & Binomial Distribution T4 ( )P 2 1 P( 0) P( 1) P( 3) P( 4) 19 11 3 3 1 112 28 28 112 39 17 1 56 56 X X X X X= = − = − = − = − = = − + + + = − = Alternatively, x 0 1 2 3 4 ( )P Xx= 24 4 1 3 1 7 2 7 2 19 112    +       = ( ) ( ) 3 3 2 4(2 3 1) 11 28 + + = 24 4 2 4 1 3 1 7 2 7 2 17 56 C   +       = 4 4 1 31 72 3 28 C   = 4 31 72 3 112   = Question 8(v) When 3n= , ( )P 2 | 3XX ( ) ( ) P 2 3 P3 X X =  ( ) ( ) P3 1 P 4 3 28 31 112 X X == −= = − 12 109= 5(i) 3 2 1 1P( 3) 4 3 2 4X    = = =       (ii) X = Number of tries John takes to open the door = { 1, 2, 3, 4} x 1 2 3 4 P(X = x) (iii) 1 1 1 1E( ) 1 2 3 4 2.54 4 4 4X =  +  +  +  = 2 2 2 2 2 1 1 1 1E( ) 1 2 3 4 7.54 4 4 4X =  +  +  +  = ( ) ( ) 222Var( ) E E 7.5 (2.5) 1.25X X X= − = − = (iv) 2P( 8)X  1 1 1P( 8) P( 1) P( 2) 4 4 2X X X=  = = + = = + = 4 1 4 1 3 1 4 3 = 4 1 4 1 2 1 3 2 4 3 =

River Valley High School, Mathematics Department 2023 Discrete Random Variable & Binomial Distribution T5 6 Given ( )P 1 0.04X == For 2,3,4,5,6r = , ( ) 1 0.04P 0.192 . 5Xr −= = = all E( ) P( ) 4.73248 t T t T t=  = = = 4.73 ( 3 sf) 22 all E( ) ( ) 23.87104 t T t P T t=  = = ( ) 222Var( ) E( ) E 23.87104 (4.73248) 1.47467305 1.47T T T= − = − = = (3 sf) t Possible Outcomes (x1, x2) Probability P(T = t ) 1 (1, 1) 0.04(0.04) 0.0016= 2 (2, 2), (1, 2), (2, 1) ( )( )0.192(0.192) 2 0.04 0.192 0.052224+= [Shown] 3 (3, 3), (1, 3), (3, 1), (2, 3), (3, 2) ( ) ( )( ) ( )0.192 (0.192) 2 0.04 0.192 2 0.192 (0.192) 0.125952 ++ = 4 (4, 4), (1, 4), (4, 1), (2, 4), (4, 2), (3, 4), (4, 3) ( ) ( )( ) ( )0.192 (0.192) 2 0.04 0.192 4 0.192 (0.192) 0.19968 ++ = 5 (5, 5), (1, 5), (5, 1), (2, 5), (5, 2), (3, 5), (5, 3), (4, 5), (5, 4) ( ) ( )( ) ( )0.192 (0.192) 2 0.04 0.192 6 0.192 (0.192) 0.273408 ++ = 6 (6, 6), (1, 6), (6, 1), (2, 6), (6, 2), (3, 6), (6, 3), (4, 6), (6, 4), (6, 5), (5, 6) ( ) ( )( ) ( )0.192 (0.192) 2 0.04 0.192 8 0.192 (0.192) 0.347136 ++ = 7 i X = Total score of the two dice = {2, 3, 4, 5, 6, 7, 8} X Possible Outcomes Probability P(X = x) 2 (1, 1) 1 1 1 4 4 16= 3 (1, 2), (2, 1) 1 1 2 1 24 4 16 8  = = 4 (2, 2), (1, 3), (3, 1) 1 1 3 34 4 16  = 5 (2, 3), (3, 2), (1, 4), (4, 1) 1 1 4 144 4 16 4  = =

River Valley High School, Mathematics Department 2023 Discrete Random Variable & Binomial Distribution T6 6 (3, 3), (2, 4), (4, 2) 1 1 3 34 4 16  = 7 (3, 4), (4, 3) 1 1 2 1 24 4 16 8  = = 8 (4, 4) 1 1 1 4 4 16= The probability distribution of X is x 2 3 4 5 6 7 8 P(X = x) 1 16 1 8 3 16 1 4 3 16 1 8 1 16 1 1 3 1 3 1 1E( ) 2 3 4 5 6 7 816 8 16 4 16 8 16X =  +  +  +  +  +  +  = 5 ii Let W = Gain of Harry in the particular game. w 36 − 3 x 6 ≠ 6 P(W = w) 3 16 13 16 In this game, E(W) = 36 3 16 – 3 13 16 = $ 69 16 or $4.3125 8(i) The probability that a salesman is successful in closing a deal with each customer is assumed to be constant. (ii) Assumption: Deals closed are independent of one another Deals closed may not be independent of one another as customers may collaborate to buy cars as a group for better bargaining power.

River Valley High School, Mathematics Department 2023 Discrete Random Variable & Binomial Distribution T7 (iii) ( ) 3060 30 30 2 P( 30) 0.03014 1 0.03014 0.2399991029 0 0.6 or 0.3999955 Since 0.5, 0.4 (1 d.p) C C p p pp p p p ==   − = − + = =  = 9(i) • Whether a call made by a