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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(A’)=1/2,p(B)=3/5,p( B’)=2/5 P(neither of them solves the problem)=p(A’)p(B’)=(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(A’)=1/2,p(B)=3/5,p( B’)=2/5 P(neither of them solves the problem)=p(A’)p(B’)=(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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1. What is probability and how is it relevant to the CA Foundation exam? |
2. What are the basic principles of probability? |
3. How can I calculate the probability of an event? |
4. What is the difference between independent and dependent events in probability? |
5. Can you provide an example of a probability calculation in the context of the CA Foundation exam? |
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