MHT CET · Maths · Probability
A coin is tossed three times. If \(\mathrm{X}\) denote the absolute difference between the number of heads and the number of tails, then, \(\mathrm{P}(\mathrm{X}=1)=\)
- A \(\frac{1}{6}\)
- B \(\frac{1}{2}\)
- C \(\frac{2}{3}\)
- D \(\frac{3}{4}\)
Answer & Solution
Correct Answer
(D) \(\frac{3}{4}\)
Step-by-step Solution
Detailed explanation
A coin is tossed 3 times
\(
\Rightarrow \mathrm{n}(\mathrm{S})=8
\)
Possibilities are : \((1 \mathrm{H}, 2 \mathrm{~T}),(2 \mathrm{H}, 1 \mathrm{~T}),(3 \mathrm{H}, 0 \mathrm{~T}),(0 \mathrm{H}, 3 \mathrm{~T})\)
Thus values of \(\mathrm{X}\) can be 1 and 3 .
Now \((1 \mathrm{H}, 2 \mathrm{~T})\) can occur in 3 ways. Also \((2 \mathrm{H}, 1 \mathrm{~T})\) can occur in 3 ways.
\(
\therefore \mathrm{P}(\mathrm{X}=1)=\frac{6}{8}=\frac{3}{4}
\)
\(
\Rightarrow \mathrm{n}(\mathrm{S})=8
\)
Possibilities are : \((1 \mathrm{H}, 2 \mathrm{~T}),(2 \mathrm{H}, 1 \mathrm{~T}),(3 \mathrm{H}, 0 \mathrm{~T}),(0 \mathrm{H}, 3 \mathrm{~T})\)
Thus values of \(\mathrm{X}\) can be 1 and 3 .
Now \((1 \mathrm{H}, 2 \mathrm{~T})\) can occur in 3 ways. Also \((2 \mathrm{H}, 1 \mathrm{~T})\) can occur in 3 ways.
\(
\therefore \mathrm{P}(\mathrm{X}=1)=\frac{6}{8}=\frac{3}{4}
\)
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