Latest Elitmus Aptitude Question SOLUTION: if 3 dice are tossed then what is the probability of getting atleast one six ?> plz tell me answer ???????????

1-((5/6)*(5/6)*(5/6))

six other other + six six other + six six six
(1/6)(5/6)(5/6)*3!/2! + (1/6)(1/6)(5/6)*3!/2! + (1/6)(1/6)(1/6)*3!/3!
=91/216

Atleast 1 six means: it may be one six or two six or all three six :

when exactly one dice(D1) occupy 1 six, then other should not equals to six so possibility=(5*5=25)
(6 1 1)...(6 1 5) = total 5 set
(6 2 1)...(6 2 5) = total 5 set
(6 3 1)...(6 3 5) = total 5 set
(6 4 1)...(6 4 5) = total 5 set
(6 5 1)...(6 5 5) = total 5 set, so clear total 25 set with only one six.its all for 1st dice(D1) here 2nd dice(D2) or 3rd dice(D3) may hold one six then total possibility =25+25+25 =75

now lets consider two six on (D1 D2)
(6 6 1)..(6 6 5)= 5 set. It may happen for (D1 D3) or (D2 D3) so tatal probability]= 5+5+5=15

lets consider all three dice hold six, possibility =1

so total evevnt = 75+15+1=91
bcz total sample =6*6*6=216

Probability = 91/216
so total 25+5+1=31

possible favorable outcomes for at least one six are 7
0 0 6
0 6 0
6 0 0
0 6 6
6 0 6
6 6 0
6 6 6
probability = 7/216

Three cases arise:

Case 1: When only one dice shows up a six

This dice can be any of the 1st, 2nd or 3rd dice. Find the probability for these three independent events and add them up to get the total probability

Probability that only 1st dice shows up a six: (probability that first dice shows up a 6) and (probability that second dice shows up other than 6) and (probability that third dice shows up other than 6)

=(1/6)*(5/6)*(5/6)

=25/216

similarly probability that 2nd dice shows up a six: (5/6)*(1/6)*(5/6) = 25/216

And, probability that 3rd dice shows up a six: (5/6)*(5/6)*(1/6) = 25/216

So probability that only one dice shows up a six: (25/216)+(25/216) +(25/216) = 75/216

Case 2:When two dice show up a six

Total number of ways of selecting a pair of dice that show up a six from a set of 3 dice are: 3C2=3

Find the probability of getting six on a pair of dice and multiply it by total number of such possible pairs

Probability of getting a six on a pair of dice = (1/6)*(1/6)*(5/6) = 5/216

So, total probability = 3*(5/216) = 15/216

Case 3: When all dice show up a six

In this case total probability is just (1/6)*(1/6)*(1/6) = 1/216



So total probability of getting at least one six = (75/216) + (15/216) + (1/216) = 91/216

Three cases arise:

Case 1: When only one dice shows up a six

This dice can be any of the 1st, 2nd or 3rd dice. Find the probability for these three independent events and add them up to get the total probability

Probability that only 1st dice shows up a six: (probability that first dice shows up a 6) and (probability that second dice shows up other than 6) and (probability that third dice shows up other than 6)

=(1/6)*(5/6)*(5/6)

=25/216

similarly probability that 2nd dice shows up a six: (5/6)*(1/6)*(5/6) = 25/216

And, probability that 3rd dice shows up a six: (5/6)*(5/6)*(1/6) = 25/216

So probability that only one dice shows up a six: (25/216)+(25/216) +(25/216) = 75/216

Case 2:When two dice show up a six

Total number of ways of selecting a pair of dice that show up a six from a set of 3 dice are: 3C2=3

Find the probability of getting six on a pair of dice and multiply it by total number of such possible pairs

Probability of getting a six on a pair of dice = (1/6)*(1/6)*(5/6) = 5/216

So, total probability = 3*(5/216) = 15/216

Case 3: When all dice show up a six

In this case total probability is just (1/6)*(1/6)*(1/6) = 1/216



So total probability of getting at least one six = (75/216) + (15/216) + (1/216) = 91/216

d) 91/216

(3C1+3C2+3C3) /6*6*6==7/216

In the first throw there can be six or other no. so, for atleast condition
1/6*5/6*5/6*3(first throw result in a six and other does not and this can be arranged in 3 ways)+1/6*1/6*5/6*3(case in which two throw result in a six and their possible arrangement)+1/6*1/6*1/6(case in which all throw result in a six)

why its not 31/216 can any1 explain me plzzz