NDA 2018
Exam NDA
Year 2018
Subject Mathematics
Chapter Probability NDA 2018 · Mathematics · Probability
If two dice are thrown and at least one of the dice shows 5, then...
If two dice are thrown and at least one of the dice shows 5, then the probability that the sum is 10 or more is
Options A 1/6 B 4/11 C 3/11 D 2/11
Step-by-step solution
A = Event of showing 5 and atleast one dice = ( 1 , 5 ) , ( 2 , 5 ) , ( 3 , 5 ) , ( 4 , 5 ) ( 5 , 5 ) , ( 6 , 5 ) , ( 5 , 1 ) , ( 5 , 2 ) , ( 5 , 3 ) , ( 5 , 4 ) , ( 5 , 6 ) n ( A ) = 11 and n ( s ) = 6 × 6 = 36 B = Event of showing sum 10 or more when atleast one dice shows 5 = ( 5 , 5 ) , ( 5 , 6 ) , ( 6 , 5 ) n ( B ) = 3 ⇒ n ( A ∩ B ) = 3 Now, P ( B A ) = P ( A ∩ B ) P ( A ) = n ( A ∩ B ) n ( S ) n ( A ) n ( S ) = 3 36 11 36 = 3 11 \begin{aligned}
& \mathrm{A}=\text{ Event\text{ of } showing } 5 \text{ and atleast one dice } \\
& =(1,5),(2,5),(3,5),(4,5)(5,5),(6,5),(5,1) \text{, } \\
& (5,2),(5,3),(5,4),(5,6) \\
& n(\mathrm{ A})=11 \\
& \text{ and } n(s)=6 \times 6=36 \\
& B=\text{ Event of showing sum } 10 \text{ or more when atleast one } \\
& \text{ dice shows } 5 \\
& =(5,5),(5,6),(6,5) \\
& n(\mathrm{ B})=3 \\
& \Rightarrow \quad n(\mathrm{ A} \cap \mathrm{ B})=3 \\
& \text{ Now, } \mathrm{P}\left(\frac{\mathrm{ B}}{\mathrm{ A}}\right)=\frac{\mathrm{P}(\mathrm{ A} \cap \mathrm{ B})}{\mathrm{P}(\mathrm{ A})} \\
& =\frac{\frac{n(\mathrm{ A} \cap \mathrm{ B})}{n(\mathrm{ S})}}{\frac{n(\mathrm{ A})}{n(\mathrm{ S})}}=\frac{\frac{3}{36}}{\frac{11}{36}}=\frac{3}{11}
\end{aligned} A = Event of showing 5 and atleast one dice = ( 1 , 5 ) , ( 2 , 5 ) , ( 3 , 5 ) , ( 4 , 5 ) ( 5 , 5 ) , ( 6 , 5 ) , ( 5 , 1 ) , ( 5 , 2 ) , ( 5 , 3 ) , ( 5 , 4 ) , ( 5 , 6 ) n ( A ) = 11 and n ( s ) = 6 × 6 = 36 B = Event of showing sum 10 or more when atleast one dice shows 5 = ( 5 , 5 ) , ( 5 , 6 ) , ( 6 , 5 ) n ( B ) = 3 ⇒ n ( A ∩ B ) = 3 Now, P ( A B ) = P ( A ) P ( A ∩ B ) = n ( S ) n ( A ) n ( S ) n ( A ∩ B ) = 36 11 36 3 = 11 3
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