MCQ - Probability | Quantitative Aptitude for CA Foundation PDF Download

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(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) 
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(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) 
 
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(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) 
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(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) 
(a)1/2  
(b)1/4  
(c)1/6  
(d)1 
Answer:c 
Solution : 
There are 2 vowels and 2 
consonants 
P(2 are vowels )=1/6 
Answer: (c)