Probaility of rolling 4 dice (1 Viewer)

[darlking]

Question- If 4 dice are thrown, find the probability that the dice will have a) only one 6 and b) at least one 6.

 

[michaeljennings]

a) this 6 can be rolled on any of the 4 turns so if you consider the 6 being rolled on the first go 1/6 X (5/6)^3, on the 2nd (5/6) x (1/6) x (5/6)^2, on the third (5/6)^2 x (1/6) x (5/6) and on the 4th (5/6)^3 x (1/6)

you should get a final answer of 4 x [(1/6)x(5/6)^3)] (when you add the above 4 cases together)

b) At least one 6 means 1 - P(no sixes)

1 - (5/6)^4

 

[darlking]

a) this 6 can be rolled on any of the 4 turns so if you consider the 6 being rolled on the first go 1/6 X (5/6)^3, on the 2nd (5/6) x (1/6) x (5/6)^2, on the third (5/6)^2 x (1/6) x (5/6) and on the 4th (5/6)^3 x (1/6)

you should get a final answer of 4 x [(1/6)x(5/6)^3)] (when you add the above 4 cases together)

b) At least one 6 means 1 - P(no sixes)




1 - (5/6)^4

Thanks so much! Can you explain a) again? I still am confused ):

 

[michaeljennings]

Thanks so much! Can you explain a) again? I still am confused ):

okay so if we are only to roll one 6 yea? This means we can either:
A) roll the 6 on the first go and have the remaining 3 rolls being a non-6,
B) the first roll being a non-6 and the 2nd roll being a 6 and the remaining two rolls being a non-6
C) First two rolls being a non-6 and the 3rd being the 6 and the last roll being a non-6
D) First 3 rolls being non-6s and the last roll being a 6


So there are 4 scenarios here, so you have to add the probability of each scenario to get your final answer (in this case all scenarios share the same probability)

Im not sure if youre learnt binomial probability but you can get the same answer doing 4C1 x (5/6)^3 x (1/6) but if you havent learnt this dont worry jsut ignore this line and focus on what I said above

 

[darlking]

okay so if we are only to roll one 6 yea? This means we can either:
A) roll the 6 on the first go and have the remaining 3 rolls being a non-6,
B) the first roll being a non-6 and the 2nd roll being a 6 and the remaining two rolls being a non-6
C) First two rolls being a non-6 and the 3rd being the 6 and the last roll being a non-6
D) First 3 rolls being non-6s and the last roll being a 6


So there are 4 scenarios here, so you have to add the probability of each scenario to get your final answer (in this case all scenarios share the same probability)

Im not sure if youre learnt binomial probability but you can get the same answer doing 4C1 x (5/6)^3 x (1/6) but if you havent learnt this dont worry jsut ignore this line and focus on what I said above

Oh, I totally get it now! Thanks a dozen.