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A die is tossed thrice. Find the probability of getting an odd number at least once.
Solution:
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Verified Answer
$\mathrm{S}=\{1,2,3,4,5,6\}, n(\mathrm{~S})=6$
Let $\mathrm{A}$ represents an odd number.
$$
\mathrm{A}=\{1,3,5\}, n(\mathrm{~A})=3 \mathrm{P}(\mathrm{A})=\frac{n(A)}{n(S)}=\frac{3}{6}=\frac{1}{2}
$$
$$
P(\bar{A})=1-P(A)=1-\frac{1}{2}=\frac{1}{2}
$$
$n=3, \mathrm{P}($ atleast one success $)=1-\mathrm{P}(\mathrm{X}=0)$
$=1-\left(\frac{1}{2}\right)^3=1-\frac{1}{8}=\frac{7}{8}$.
Let $\mathrm{A}$ represents an odd number.
$$
\mathrm{A}=\{1,3,5\}, n(\mathrm{~A})=3 \mathrm{P}(\mathrm{A})=\frac{n(A)}{n(S)}=\frac{3}{6}=\frac{1}{2}
$$
$$
P(\bar{A})=1-P(A)=1-\frac{1}{2}=\frac{1}{2}
$$
$n=3, \mathrm{P}($ atleast one success $)=1-\mathrm{P}(\mathrm{X}=0)$
$=1-\left(\frac{1}{2}\right)^3=1-\frac{1}{8}=\frac{7}{8}$.
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