ACJC H1 LCP1 Solution

1 2024 H1 Maths JC2 LCP1 Solutions Qn Solutions 1(a) A = {2,4,6} B = {3,6} C={1,2} B and C are mutually exclusive since BC = (b) A and B are independent and A and C are independent Method 1 P(A) = 31 62= P(B) = 21 63= P(C) = 21 63= {6}A B= { 1 }A C= P  1(A B)= 6 PC 1(A )= 6 1 1 1P( ) P( )= P( )2 3 6  = = A B A B 1 1 1P( ) P(C)= P( )2 3 6A A C  = =  Hence A and B are independent and A and C are independent events. Method 2 For set B only “6” is even, hence P(A|B) = 1 2 P(A) = 31 62= Hence P(A|B)=P(A).  A and B are independent. For set C only “2” is even, hence P(A|C) = 1 2 P(A) = 31 62= Hence P(A|C)=P(A).  A and C are independent.

2 2(a) P(faulty mug)= 990.35 0.03 0.65 0.06 0.04952000 +  = = b(i) P(exactly one mug faulty) = 0.0495 (1 0.0495) 2 0.0941 (to 3 sig fig) −  = b(ii) Method 1 P(exactly one faulty, given that both made by A) = 2910.03 (1 0.03) 2 0.05825000 −  = = Method 2 P(exactly one faulty, given that both made by A) P(exactly one mug is faulty both made b y ) P(both made by ) A A = 0.35 0.03 0.35 (1 0.03) 2 0.35 0.35    − =  2910.03 (1 0.03) 2 0.05825000=  −  = = 3(a) P(L|T)= 6 25 2 3 P(L T) 9 P(T) 25  == (b) P( ) = P( ) + P( ) - P( )L T L T L T = 2 2 6 62 5 3 25 75+ − = ''P( )LT = P( ( )'LT )= 1 ( )P L T− = 62 131 75 75−= = 0.173 (to 3 sf) (c) Method 1 [P( ) P( )] + [P( ) P( )]L L T T L T−  −  = 2 6 2 6( ) ( )5 25 3 25− + − = 44 75 =0.587 (to 3 sf) Method 2 P( ) P( )L T L T −  = 62 6 44 75 25 75−= =0.587

3 4(a) Consider the 2 Biology books as 1 unit Number of different arrangements = 9! 2! 725760= (b) Method 1 Consider the 3 Chemistry books as 1 unit Number of different arrangements such that the Chemistry books are together = 8! 3! Probability that the Chemistry books are not all next to one another = 8! 3! 141 0.93310! 15 − = = (to 3 sig fig) Method 2 P(Chemistry books are not all next to one another) = P(all separated) + P(2 together, 1 separated) n(all Chemistry books are separated) = 8 37! × P n(1 Chemistry book is separated from 2 Chemistry books which are together) = 83 227! × P 2!C Probability that the Chemistry books are not all next to one another = 8 8 3 3 2 27! 7! 2! 10!  +   P P C = 14 0.93315= ( to 3 sig fig) (c) Number of possible selections – Number of selections whereby none of the Chemistry is included = 10 7 44 175CC−= (d) Probability that at most one Chemistry book is selected = 7 7 3 4 3 1 10 4 C C C C + = 2 3