MHT CET · Maths · Definite Integration
A fair coin is tossed 4 times. If \(X\) a random variable which indicates number of heads, then \(\mathrm{P}[\mathrm{X} < 3]=\)
- A \(\frac{10}{16}\)
- B \(\frac{1}{16}\)
- C \(\frac{12}{16}\)
- D \(\frac{11}{16}\)
Answer & Solution
Correct Answer
(D) \(\frac{11}{16}\)
Step-by-step Solution
Detailed explanation
A coin is tossed 4 times
\(
\therefore \mathrm{n}(\mathrm{S})=2^4=16
\)
Following possibilities exist.
(i) All Heads \(\Rightarrow 1\) way
(ii) 3 Heads, 1 Tail \(\Rightarrow \frac{4 !}{3 !}=4\) ways
(iii) 2 Heads, 2 Tails \(\Rightarrow \frac{4 !}{2 ! 2 !}=6\) ways
(iv) 1 Head, 3 Tails \(\Rightarrow \frac{4 !}{3 !}=4\) ways
(v) 0 head, 4 Tails \(\Rightarrow 1\) way
\(\therefore\) Required probability.
\(=\mathrm{P}(\mathrm{x}=0,1,2)=\frac{1}{16}+\frac{4}{16}+\frac{6}{16}\)
\(=\frac{11}{16}\)
\(
\therefore \mathrm{n}(\mathrm{S})=2^4=16
\)
Following possibilities exist.
(i) All Heads \(\Rightarrow 1\) way
(ii) 3 Heads, 1 Tail \(\Rightarrow \frac{4 !}{3 !}=4\) ways
(iii) 2 Heads, 2 Tails \(\Rightarrow \frac{4 !}{2 ! 2 !}=6\) ways
(iv) 1 Head, 3 Tails \(\Rightarrow \frac{4 !}{3 !}=4\) ways
(v) 0 head, 4 Tails \(\Rightarrow 1\) way
\(\therefore\) Required probability.
\(=\mathrm{P}(\mathrm{x}=0,1,2)=\frac{1}{16}+\frac{4}{16}+\frac{6}{16}\)
\(=\frac{11}{16}\)
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