There are four cards bearing numbers 1, 3, 5, 7 in bag A. There are three cards bearing numbers 2, 4, 6 in bag B. A card is drawn at random first from bag A and then from bag B alternately. Find the probability that the sum of these two cards drawn is (i) seven (ii) multiple of 3 (iii) more than 7.
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Answer:
S= 1,2 1,4 1,6
3,2 3,4 3,6
5,2 , 5,4 ; 5,6 ; 7,2 ; 7,4 ; 7,6
n(S) = 12
Step-by-step explanation:
i) A= 5,2 ; 3,4 ; 1,6
n(A) = 3
p(A) = 3/12 = 1/4
and so on
ii) 7/12
iii)3/4