A die with x sides has consecutive integers on its sides.If the probability of not getting a 4 on either of the 2 tosses is 36/49, how many sides does the die have?
a. 4
b. 5
c. 7
d. 8
e. 13
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naveen451
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IMO Cruplun wrote:A die with x sides has consecutive integers on its sides.If the probability of not getting a 4 on either of the 2 tosses is 36/49, how many sides does the die have?
a. 4
b. 5
c. 7
d. 8
e. 13
I AM NOT SURE BUT
explanation
if suppose there are 7 sides
then probability of getting 4 =1/7
probability of not getting 4=1-1/7 =6/7
probability pf not getting 4 for two times =6/7*6/7=36/49
thus die has 7 sides.
if i am wrong plz correct me
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tauseef10jan2004
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let the die has n sides
then probability of getting 4 = 1/n
probability of not getting 4 = 1-(1/n) = (n-1)/n
probability of not getting 4 twice = ((n-1)/n)*((n-1)/n) = sqr((n-1)/n) = 36/49
i.e. (n-1)/n = 6/7
on solving this n = 7
answer is c. 7
then probability of getting 4 = 1/n
probability of not getting 4 = 1-(1/n) = (n-1)/n
probability of not getting 4 twice = ((n-1)/n)*((n-1)/n) = sqr((n-1)/n) = 36/49
i.e. (n-1)/n = 6/7
on solving this n = 7
answer is c. 7
I just saw 7^2 in the denominator and chose C, knowing that in most of the probability-of-die-rolls questions I have faced, the solution often had 6^n in the denominator (n = the number of rolls of a standard dice).
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shashank.ism
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Its quite simple. Let there be n sides of die.
so probability of getting 4 in one throw = 1/n
and probability of not getting 4 in one throw = (n-1)/n
so probability of not getting a 4 i n both the 2 throw = {(n-1)/n}^2
--> {(n-1)/n}^2 = 36/49
--> {(n-1)/n} = 6/7
--> n = 7.
Hence Ans C
so probability of getting 4 in one throw = 1/n
and probability of not getting 4 in one throw = (n-1)/n
so probability of not getting a 4 i n both the 2 throw = {(n-1)/n}^2
--> {(n-1)/n}^2 = 36/49
--> {(n-1)/n} = 6/7
--> n = 7.
Hence Ans C
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