Page 1 CPT Section D, Quantitative Aptitude, Chapter 13 Prof(Dr.) P .R.Vittal Page 2 CPT Section D, Quantitative Aptitude, Chapter 13 Prof(Dr.) P .R.Vittal Page 3 CPT Section D, Quantitative Aptitude, Chapter 13 Prof(Dr.) P .R.Vittal (a)2/9 (b)1/4 (c)4/9 (d)1/3 Answer: a Answer : Sample space S={[1,3],[2,4],[3,5],[4,6],[3,1],[4,2],[5,3], [6,4]} P(difference is 2 )=8/36 = 2/9 Answer : (a) Page 4 CPT Section D, Quantitative Aptitude, Chapter 13 Prof(Dr.) P .R.Vittal (a)2/9 (b)1/4 (c)4/9 (d)1/3 Answer: a Answer : Sample space S={[1,3],[2,4],[3,5],[4,6],[3,1],[4,2],[5,3], [6,4]} P(difference is 2 )=8/36 = 2/9 Answer : (a) (a)3/10 (b)2 /5 (c)1/5 (d)1/4 Answer:c Solution :p(A)=1/2,p(Aâ€™)=1/2,p(B)=3/5,p( Bâ€™)=2/5 P(neither of them solves the problem)=p(Aâ€™)p(Bâ€™)=(1/2).(2/5) =1/5 Answer: (c) Page 5 CPT Section D, Quantitative Aptitude, Chapter 13 Prof(Dr.) P .R.Vittal (a)2/9 (b)1/4 (c)4/9 (d)1/3 Answer: a Answer : Sample space S={[1,3],[2,4],[3,5],[4,6],[3,1],[4,2],[5,3], [6,4]} P(difference is 2 )=8/36 = 2/9 Answer : (a) (a)3/10 (b)2 /5 (c)1/5 (d)1/4 Answer:c Solution :p(A)=1/2,p(Aâ€™)=1/2,p(B)=3/5,p( Bâ€™)=2/5 P(neither of them solves the problem)=p(Aâ€™)p(Bâ€™)=(1/2).(2/5) =1/5 Answer: (c) a)2 /5 (b)1/5 (c)3/5 (d)3/11 Answer: d Solution :Since the total probability is 1, p(A)+p(B)+p(C)=1. Hence we have 2p(B)+p(B)+(2/3)(p(B)=1 P(B)=3/11 Answer: (d)Read More

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