Latest self Aptitude Question SOLUTION: Ina single throw of two dice, what is the probability that neither a doublet nor a total of 8 will appear? a)7/15 b)5/18 c)13/18 d)3/16

• Placement Papers
• /
• self

self Question Papers:

(#M40023901) SELF question PROBABILITY Keep an EYE

Ina single throw of two dice, what is the probability that neither a doublet nor a total of 8 will appear?
a)7/15
b)5/18
c)13/18
d)3/16

(10 years ago)

Unsolved Read Solution (4)

Is this Puzzle helpful?   (3)   (4) Submit Your Solution

Read Solution (4) :

sreebhuvan  (10 years ago)

doublets (1,1)(2,2)...(4,4),..(6,6)=> n(A)=6

ans: 13/18
total 8=> (2,6)(3,5) (4,4) (5,3)(6,2)=>n(B)=5
A I B=>(4,4) n(A I B)=1
n(A U B)=n(A)+n(B)-n(A I B)=6+5-1=10
due to neither a doublet nor a total of 8
p=(36-10)/36=26/36
=13/18

Is this solution Helpfull? Yes (5) | No   

Bholanath   (10 years ago)

An. c) 13/18
doublet {(1,1),(2,2),(3,3),(4,4),(5,5),(6,6)}
total 8 will be from the combination of {(2,6),(6,2),(3,5),(5,3),(4,4)}

total no. of combination 6+5-1=10;(-1 because (4,4) taken two times)

so, no. of possibilities that neither a doublet nor a total of 8 will be
=36-10=26
total no. possibilities =6*6=36
so,
probability = 26/36=13/18

Is this solution Helpfull? Yes (3) | No   

md athar qamar  (10 years ago)

let pa(A) be probability of doublet & p(B) be the probability of total of 8.now we need to find p(A^c intersection B^c).
now p(A^c intersection B^c)=p(AUB)^c=1-p(AUB)=1-p(A)-p(B)+p(A intersection B)
again,p(A)=6/36, p(B)=5/36,p(A intersection B)=1/36.
putting all terms in above equation we get=13/18 ans.

Is this solution Helpfull? Yes (2) | No   

MUKESH KUMAR CHAURASIA  (10 years ago)

(11,22,33,44,55,66,26,35,53,62)
10/36=5/18

Is this solution Helpfull? Yes | No (4)  

Please Login to Read Solution.