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Question

A die is rolled twice. What is the probability of getting a sum equal to 9?

2/3

2/9

1/6

1/9

None of these

ANSWER : 4

Descrption

Total number of out comes possible when a die is rolled = 6 
Hence, total number of out comes possible when a die is rolled twice, n(s) = 6 &lt;math xmlns = &quot;http://www.w3.org/1998/Math/MathML&quot;&gt;&lt;mo&gt;&amp;#xD7;&lt;/mo&gt;&lt;/math&gt; 6 = 36 
E = Geting a sum of 9 when the two dice fell = {(3,6), (4,5), (5,4), (6,3)} 
Hence n (E) = 4 
P(E) = &lt;math xmlns = &quot;http://www.w3.org/1998/Math/MathML&quot;&gt;&lt;mfrac&gt;&lt;mrow&gt;&lt;mi&gt;n&lt;/mi&gt;&lt;mo&gt;(&lt;/mo&gt;&lt;mi&gt;E&lt;/mi&gt;&lt;mo&gt;)&lt;/mo&gt;&lt;/mrow&gt;&lt;mrow&gt;&lt;mi&gt;n&lt;/mi&gt;&lt;mo&gt; 
(&lt;/mo&gt;&lt;mi&gt;S&lt;/mi&gt;&lt;mo&gt;)&lt;/mo&gt;&lt;/mrow&gt;&lt;/mfrac&gt;&lt;mo&gt; = &lt;/mo&gt;&lt;mo&gt;&amp;#xA0;&lt;/mo&gt;&lt;mfrac&gt;&lt;mn&gt;4&lt;/mn&gt;&lt;mn&gt;36&lt;/mn&gt;&lt;/mfrac&gt;&lt;mo&gt;&amp;#xA0;&lt;/mo&gt;&lt;mo&gt; = 
&lt;/mo&gt;&lt;mo&gt;&amp;#xA0;&lt;/mo&gt;&lt;mfrac&gt;&lt;mn&gt;1&lt;/mn&gt;&lt;mn&gt;9&lt;/mn&gt;&lt;/mfrac&gt;&lt;/math&gt;

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