ProbabilityMedium
Question
A bag contains $(N + 1)$ coins -N fair coins, and one coin with 'Head' on both sides. A coin is selected at random and tossed. If the probability of getting 'Head' is $\frac{9}{16}$, then N is equal to:
Options
A.5
B.7
C.8
D.9
Solution
Sol. [Key Concept: $P(Head) = \frac{N}{N + 1} \cdot \frac{1}{2} + \frac{1}{N + 1} \cdot 1$; set this equal to $\frac{9}{16}$ and solve for $N$.]
Total $(N + 1)$ coins
$N \Rightarrow$ fair coins
$1 \Rightarrow$ Head on both side
$${P\left( 1^{+} \right) = \left( \frac{N}{N + 1} \right)\left( \frac{1}{2} \right) + \left( \frac{1}{\text{ }N + 1} \right)1 }{= \frac{N + 2}{2(\text{ }N + 1)} = \frac{9}{16} }{8\text{ }N + 16 = 9\text{ }N + 9 }{\Rightarrow N = 7 }$$
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