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So i know the probability rule of addition.

Getting 2 or 5 in two throws should be P(2)+ P(5). P(2) = 1/6, P(5) = 1/6 so the combined so it should be 1/3.

I tried to visualize but not able to do so correctly. 11,12,13,14,15,16, 21,22,23,24,25,26,31,32, ....6,6 total of 36 possibilities.

12,15,21,22,23,24,25,26,31,35,42,45,51,52,53,54,55,56,61,65 out of which 20 possibilities

so the probability should be 20/36 which is not 1/3

where am i doing wrong?

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    Easier to use the complement: 1 - (4/6)(4/6) = 20/36 – 2017-01-22
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    very nice, what is your insight on when to use this kind of complement approach – 2017-01-22

3 Answers 3

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You did not take into account the fact that a 2 or a 5 could be obtained in either the first roll or the second roll. You must add these probabilities. Also, you must take into account the possibility that both events occur.

The probability of obtaining a 2 on the first roll is $1/6$. The probability of obtaining a 2 on the second roll is also $1/6$. Similarly, the probability of obtaining a 5 on the first roll is $1/6$, and the probability of obtaining a 5 on the second roll is also $1/6$. Adding those four probabilities yields $$4 \cdot \frac{1}{6} = \frac{2}{3}$$ However, we have counted those outcomes in which we obtain a 2 on both rolls, a 5 on both rolls, a 2 on the first roll and a 5 on the second roll, and a 5 on the first roll and a 2 on the second roll twice. Each of those four events has probability $1/36$. Hence, the probability of obtaining a 2 or a five in two throws of a die is $$4 \cdot \frac{1}{6} - 4 \cdot \frac{1}{36} = \frac{2}{3} - \frac{1}{9} = \frac{5}{9}$$ as you found by listing the possibilities.

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    what is the insight that needs to be developed to understand these kind of combinations when trying to just use probability rules instead of exhaustive combinations. – 2017-01-22
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Your both methods have some mistake.

Method 1 -

Getting 2 on first number and any other number on second except 5. But 2 can be on second die also. So multiply cases with 2.

$2 \left(\frac16 \times \frac46\right)$

= $\left(\frac13 \times \frac23\right)$

= $\left(\frac29\right)$

Similarly for getting 5 on first number and any other number on second except 2. But 5 can be on second die also. So multiply cases with 2.

$2 \left(\frac16 \times \frac46\right)$

= $\left(\frac13 \times \frac23\right)$

= $\left(\frac29\right)$

Case with 2 on both dice or 5 on both dice.

$\left(\frac16 \times \frac16 + \frac16 \times \frac16\right)$

= $\left(\frac1{36} + \frac1{36}\right)$

= $\left(\frac2{36}\right)$

Case with 2 on first die and 5 on second or vice versa.

$\frac16 \times \frac16 + \frac16 \times \frac16$

= $\left(\frac1{36} + \frac1{36}\right)$

= $\left(\frac2{36}\right)$

Combining these,

$\left(\frac29 + \frac29 + \frac2{36} + \frac2{36}\right)$

= $\left(\frac{20}{36}\right)$

Method 2 -

21, 22, 23, 24, 25, 26, 12, 32, 42, 52, 62, 15, 35, 45, 55, 65, 51, 53, 54, 56 = 20 cases.

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    nice, what do i need to do build this kind of thinking in coming up with these cases? – 2017-01-22
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    Second method approach to write cases then think use of formulas. – 2017-01-22
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    My method - Like either 2 on first dice, now fix it on first position and write all other numbers on second position in sequence. So we have 21, 22, 23, 24, 25, 26. Now we can have 2 on second position so fix it and write first number in sequence. So we have 12, 22, 32, 42, 52, 62. Then cut one number from duplicate numbers. – 2017-01-22
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    As 22 comes twice. So we cut one case with 22. – 2017-01-22
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I presume that if you know the probability rule for addition, you also know that for multiplication, often called the fundamental counting principle

You need either a 2, or a 5 or both.
In such situations, use of the complement can greatly simplifies computations.

The outcome we don't want, P(neither $2$ nor $5$),

thus $P(2\;or\;5) = 1 - \dfrac46\cdot\dfrac46 = \dfrac{20}{36}= \dfrac59$