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September 2, 2014

September 2, 2014

Posted by **Pam** on Sunday, December 3, 2006 at 5:23pm.

What is the probability of at least one 2 or one 5?

What is the probability of a sum of 7?

"what is the probability that you get a 2 and a 5 without regard to which is on which die? "

1/6* 1/6* 2 = 1/18

There are two possibilities both with probability 1/6 * 1/6.

"What is the probability of at least one 2 or one 5?"

There are two ways to do this. Let's first do it using an elementary trick:

Calculate the probability that you don't satisfy this critrium and then take 1 minus this probability:

NOT [at least one 2 OR one 5]

You simplify this expression by using the rule that says:

NOT[A OR B] = NOT[A] AND NOT[B]

In other words, for [A OR B] to be not true, both A and B must be false.

NOT [at least one 2 OR one 5] =

Not[at least one 2] And Not[one 5] =

= [No 2's] And [no 5's Or two 5's]

So, you can't have any 2's and you must have either no five or two five. The "No 2's" criterium means that for each dice can only have 5 possible values. To satisfy the second constraint you must either have no 5's for which there are 4*4 = 16 possibilities (both the two and the five are illegal) or you must have two fives, this can be realized in only one way.

The zero fives and the two fives are mutually exclusive so you can add up the number of ways you can realize the two criteria. So, there are 17 ways to satisfy the criterium of:

[No 2's] And [no 5's Or two 5's]

The probablity is thus 17/36 and the probability that this is not satisfied is thus:

1-17/36 = 19/36

A more direct way to calculate this is as follows. The probability of any given outcome is 1/36, so we can write the probability as the number of dice configurations that satisfy the criterium divided by 36.

Suppose you can calculate easily thenumber of caonfigurations such that criterium A is satisfied and you can also calculate the number of configuraions such that some other criterium B is satisfied. Then, the number of ways that A or B is satisfied is given by:

N[A OR B] = N[A] + N[B] - N[A AND B]

N[A] is the number of configurations such that A is satisfied and N[B] is the number of configurations such that B is satisfied. Suppose that it is possible to satisfy both. Then, any configuration that satisfies both A and B will be counted both in N[A] and in

N[B]. That's a double counting and to correct for that we must subtract

N[A AND B]

N[at least one 2 or one 5] =

N[at least one 2] + N[one 5] -

N[at least one 2 AND one 5]

N[at least one 2] can be evaluated using this rule or using by calculating the number of configurations that don't satisfy this criterium and then subrtacting 36 from that.

N[at least one 2] =

N[die one shows a 2 OR die 2 shows a 2] =

N[die one shows a 2] + N[die two shows a 2] -

N[die one shows a 2 AND die two shows a 2] =

6+6-1 =11

Alternatively you can calculate the number of configurations in which you don't have at least one 2. Then you have 5 possiblities for both dice so there are 25 possibilites in total. And 36 - 25 = 11

So we have found that:

N[at least one 2] = 11

THe next term we must evaluate is:

N[one 5]. This is easy. To have one 5 one die must show a five and the other must swow something else. If the first die shows the five there are 5 possibilites for the other die. And if the second one shows the 5 there are 5 possibilites for the first die. So, there are a total of ten possibilities:

N[one 5] = 10

Finally we must evaluate:

N[at least one 2 AND one 5].

This is easy, because if one of the dice is 5 and you must have at least one die to show a 2, the other die must be a two. You then have two possibilities one for die one showing a two and the other one showing a five and vice versa. So, we have:

N[at least one 2 AND one 5]= 2

The answer is thus:

N[at least one 2 or one 5] =

N[at least one 2] + N[one 5] -

N[at least one 2 AND one 5]

11 + 10 - 2 = 19

and the probablity is 19/36 in agreement with what we found above.

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