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from two throws of a die, find the probability of getting two sixes
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And totally wrong. On 1/29/15, kahching wrote: > On 1/29/15, Probability wrote: >> from two throws of a die, find the probability of >> getting two sixes > > The way I am understanding your question: Throw one die, pick > it up, and throw it again. That would be 1 in 6 on the first > throw and then 1 in 6 on the second throw. Add, simplify,...See More
Jan 29, 2015
BT On 1/29/15, kahching wrote: > On 1/29/15, Probability wrote: >> from two throws of a die, find the probability of >> getting two sixes > > The way I am understanding your question: Throw one die, pick > it up, and throw it again. That would be 1 in 6 on the first > throw and then 1 in 6 on the second throw. Add, simplify,...See More
Jan 30, 2015
More importantly On 1/30/15, BT wrote: > On 1/29/15, kahching wrote: >> On 1/29/15, Probability wrote: >>> from two throws of a die, find the probability of >>> getting two sixes >> >> The way I am understanding your question: Throw one die, > pick >> it up, and throw it again. That would be 1 in 6 on the first >&g...See More
Jan 30, 2015
ok Make a sample space

1,1 1,2 1,3 1,4 1,5 1,6 2,1 2,2 2,3 2,4 2,5 2,6 3,1 3,2
3,3 3,4 3,5 3,6 4,1 4,2 4,3 4,4 4,5 4,6 5,1 5,2 5,3 5,4
5,5 5,6 6,1 6,2 6,3 6,4 6,5 6,6

P(two 6s) = 1/36

as shown by: 1/6(first roll) x 1/6(second roll) = 1/36

Look what else you can see:

P(doubles) = 6/36 = 1/6 P (one...See More
Feb 24, 2015
ok On 2/24/15, ok wrote:

> Look what else you can see: > > P(doubles) = 6/36 = 1/6 > P (one 4) = 11/26 > > etc.

Whoops, I counted the two fours as a favorable event. And I turned a 36 into a 26. So 10/36
Feb 24, 2015


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