Probability die tossed

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Zelda22

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A symmetric die is tossed three times. Find the probability that all three tosses yield the same result

I'm a bit confused. The possible outcome each time is 6, so 6*6*6= 216 possible outcomes

A={ (1,1,1),(2,2,2),(3,3,3),(4,4,4,),(5,5,5),(6,6,6)}

p= p(1)+p(2)+p(3)+p(4)+p(5)+p(6)
= 1/216 * 6 = 6/216 = 1/36

Is this correct? Thanks
 
Zelda22 said:
A symmetric die is tossed three times. Find the probability that all three tosses yield the same result

I'm a bit confused. The possible outcome each time is 6, so 6*6*6= 216 possible outcomes

A={ (1,1,1),(2,2,2),(3,3,3),(4,4,4,),(5,5,5),(6,6,6)}

p= p(1)+p(2)+p(3)+p(4)+p(5)+p(6)
= 1/216 * 6 = 6/216 = 1/36

Is this correct? Thanks
Click to expand...
Hi Zelda,
I got the same answer. What are you confused about?
:)
 
Zelda22 said:
I wasn't sure if I should add the outcomes for each number or not.
Click to expand...
This is how I remember when to add or multiply. When you see "and" you multiply, and when you see "or" you add. So with your problem, you're looking for the probability of (1,1,1) or (2,2,2) or (3,3,3) or...(6,6,6). So you would add p(1,1,1) + p(2,2,2) +... + p(6,6,6).
On the other hand, when you're calculating the probability of getting 1 and 1 and 1, you multiply (1/6)*(1/6)*(1/6).
Hope this helps.
:)
 
AvgStudent said:
This is how I remember when to add or multiply. When you see "and" you multiply, and when you see "or" you add. So with your problem, you're looking for the probability of (1,1,1) or (2,2,2) or (3,3,3) or...(6,6,6). So you would add p(1,1,1) + p(2,2,2) +... + p(6,6,6).
On the other hand, when you're calculating the probability of getting 1 and 1 and 1, you multiply (1/6)*(1/6)*(1/6).
Hope this helps.
:)
Click to expand...
... but it is worth remembering that multiplication only works for independent events, and addition only for non-intersecting ones.
 
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