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kakasveta [241]
2 years ago
12

you roll two dice what is the probability that the sum of the dice is less than 5 and one dice shows a 2? ...?

Mathematics
1 answer:
astraxan [27]2 years ago
3 0
<span>Lets say the 1st die rolled a 2 - there would be 2 combinations for which the sum of dice being < 5 : 2,1 2,2 Now say the 2nd die rolled a 2 - there would be 2 combinations for which the sum of dice being < 5 : 1,2 2,2 Now we want to count all cases where either dice showed a 2 and sum of the dice was < 5. However note above that the roll (2,2) is counted twice. So there are three unique dice roll combinations which answer the criteria of at least one die showing 2, and sum of dice < 5: 1,2 2,1 2,2 The total number of unique outcomes for two dice is 6*6=36 . So, the probability you are looking for is 3/36 = 1/12</span>
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Step by step solution :<span>Step  1  :</span><span>Equation at the end of step  1  :</span><span><span> ((y2) • (q - 4)) - c • (q - 4) </span><span> Step  2  :</span></span><span>Equation at the end of step  2  :</span><span> y2 • (q - 4) - c • (q - 4) </span><span>Step  3  :</span>Pulling out like terms :

<span> 3.1 </span>     Pull out     q-4 

After pulling out, we are left with : 
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Trying to factor as a Difference of Squares :

<span> 3.2 </span>     Factoring: <span> y2-c</span> 

Theory : A difference of two perfect squares, <span> A2 - B2  </span>can be factored into <span> (A+B) • (A-B)

</span>Proof :<span>  (A+B) • (A-B) =
         A2 - AB + BA - B2 =
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        <span> A2 - B2</span>

</span>Note : <span> <span>AB = BA </span></span>is the commutative property of multiplication. 

Note : <span> <span>- AB + AB </span></span>equals zero and is therefore eliminated from the expression.

Check : <span> y2  </span>is the square of <span> y1 </span>

Check :<span> <span> c1  </span> is not a square !! 
</span>Ruling : Binomial can not be factored as the difference of two perfect squares

Final result :<span> (q - 4) • (y2 - c) </span><span>
</span>
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