A random experiment has five outcomes w_1, w_2, w_3, w_4 and w_5 . The probabilities of the occurrence of the outcomes w_1, w_2, w_3, w_4 and w_5 are respectively 1/6, a, b and 1/12 such that 12 a+12 ~b-1=0 . Then the probabilities of occurrence of the outcome w_3 is

Correct answer: (A) 2/3. p (w_1 )=1/6 p (w_2 )=a ⇒ 1/6+a+b+x+1/12=1 p (w_3 )=b ⇒ 12(a+b+x)=9 p (w_4 )=c ⇒ a+b+x=3/4 p (w_5 )=1/12 ⇒ 1/12+x=3/4 ⇒ x=2/3

Q68 / 180 · KCET 2025 · Mathematics

A random experiment has five outcomes $\mathrm{w}_1, \mathrm{w}_2, \mathrm{w}_3, \mathrm{w}_4$ and $\mathrm{w}_5$. The probabilities of the occurrence of the outcomes $\mathrm{w}_1, \mathrm{w}_2, \mathrm{w}_3, \mathrm{w}_4$ and $\mathrm{w}_5$ are respectively $\frac{1}{6}, \mathrm{a}, \mathrm{b}$ and $\frac{1}{12}$ such that $12 \mathrm{a}+12 \mathrm{~b}-1=0$. Then the probabilities of occurrence of the outcome $w_3$ is

Correct answer
A
$\frac{2}{3}$
B
$\frac{1}{3}$
C
$\frac{1}{6}$
D
$\frac{1}{12}$

Solution

$\begin{array}{ll}\mathrm{p}\left(\mathrm{w}_1\right)=\frac{1}{6} & \\ \mathrm{p}\left(\mathrm{w}_2\right)=\mathrm{a} & \Rightarrow \frac{1}{6}+\mathrm{a}+\mathrm{b}+\mathrm{x}+\frac{1}{12}=1 \\ \mathrm{p}\left(\mathrm{w}_3\right)=\mathrm{b} & \Rightarrow 12(\mathrm{a}+\mathrm{b}+\mathrm{x})=9 \\ \mathrm{p}\left(\mathrm{w}_4\right)=\mathrm{c} & \Rightarrow \mathrm{a}+\mathrm{b}+\mathrm{x}=\frac{3}{4} \\ \mathrm{p}\left(\mathrm{w}_5\right)=\frac{1}{12} & \Rightarrow \frac{1}{12}+\mathrm{x}=\frac{3}{4} \Rightarrow \mathrm{x}=\frac{2}{3}\end{array}$

Hence, the correct answer is option: 'A'.

Solution

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