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NDA 2012
Exam
NDA
Year
2012
Subject
Mathematics
Chapter
Probability

NDA 2012 · Mathematics · Probability

The dice each numbered from 1 to 6 are thrown together. Let A and B...

The dice each numbered from 1 to 6 are thrown together. Let A and B be two events given by A: even number on the first die B: number on the second die is greater than 4 What is P(A∪B)\mathrm{P}(\mathrm{A} \cup \mathrm{B}) equal to?

Options

  1. A12\frac{1}{2}
  2. B14\frac{1}{4}
  3. C23\frac{2}{3}
  4. D16\frac{1}{6}

Correct answer

C. 23\frac{2}{3}

Step-by-step solution

 A: even numbers ={2,4,6}B: number >4={5,6}A∩B={6}\begin{aligned} & \text{ A: even numbers }=\{2,4,6\} \\ & \mathrm{B}: \text{ number } > 4=\{5,6\} \mathrm{A} \cap \mathrm{B}=\{6\}\end{aligned}P(A∩B)=16P(A)=36=12\begin{aligned} \mathrm{P}(\mathrm{A} \cap \mathrm{B}) & =\frac{1}{6} \\ \mathrm{P}(\mathrm{A}) & =\frac{3}{6}=\frac{1}{2}\end{aligned}P(B)=26=13P(A∪B)=P(A)+P(B)−P(A∩B)\begin{aligned} P(B) & =\frac{2}{6}=\frac{1}{3} \\ P(A \cup B) & =P(A)+P(B)-P(A \cap B)\end{aligned}=12+13−16=3+2−16=46=23.\begin{aligned} & =\frac{1}{2}+\frac{1}{3}-\frac{1}{6}=\frac{3+2-1}{6} \\ & =\frac{4}{6}=\frac{2}{3}.\end{aligned}$
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