AP EAMCET · Maths · Probability
When six coins are tossed simultaneously, the probability of getting at least 4 heads is
- A \(\frac{11}{64}\)
- B \(\frac{15}{64}\)
- C \(\frac{11}{32}\)
- D \(\frac{15}{32}\)
Answer & Solution
Correct Answer
(C) \(\frac{11}{32}\)
Step-by-step Solution
Detailed explanation
Total possible chances \(=2^6=64\) Required probability \[ \begin{gathered} =P(\text { Getting } 4 \text { Heads })+P(\text { Getting } 5 \text { Heads }) \\ \{\text { TTHHHH }\} \quad\{\text { THHHHH }\} \\ +P(\text { Getting } 6 \text { Heads }) \end{gathered} \] \{Нннннн \}…
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