Let $N$ be the number of fair coins. The total number of coins in the bag is $(N + 1)$.
Let $F$ be the event of selecting a fair coin, and $B$ be the event of selecting the biased coin. Let $H$ be the event of getting a Head.
The total probability of getting a Head, $P(H)$, is given by the law of total probability:
$P(H) = P(H|F) \times P(F) + P(H|B) \times P(B)$
We are given that $P(H) = \frac{9}{16}$. Substituting the probabilities:
$\frac{9}{16} = \left(\frac{1}{2} \times \frac{N}{N + 1}\right) + \left(1 \times \frac{1}{N + 1}\right)$
Simplify the equation:
$\frac{9}{16} = \frac{N}{2(N + 1)} + \frac{1}{N + 1}$
Combine the terms on the right side by finding a common denominator:
$\frac{9}{16} = \frac{N}{2(N + 1)} + \frac{2}{2(N + 1)}$
$\frac{9}{16} = \frac{N + 2}{2(N + 1)}$
Now, cross-multiply:
$9 \times 2(N + 1) = 16 \times (N + 2)$
$18(N + 1) = 16(N + 2)$
$18N + 18 = 16N + 32$
Isolate the terms with $N$:
$18N - 16N = 32 - 18$
$2N = 14$
Solve for $N$:
$N = \frac{14}{2}$
$N = 7$
A bag contains 19 unbiased coins and one coin with head on both sides. One coin drawn
at random is tossed and head turns up. If the probability that the drawn coin was unbiased,
is $\frac{m}{n}$, gcd (m, n) = 1, then $n^2 - m^2$ is equal to :
Given three indentical bags each containing 10 balls, whose colours are as follows :
| Red | Blue | Green | |
| Bag I | 3 | 2 | 5 |
| Bag II | 4 | 3 | 3 |
| Bag III | 5 | 1 | 4 |
A person chooses a bag at random and takes out a ball. If the ball is Red, the probability that it is from bag I is $p$ and if the ball is Green, the probability that it is from bag III is $q$, then the value of $\left( \frac{1}{p} + \frac{1}{q} \right)$ is:
| Class | 5-10 | 10-15 | 15-20 | 20-25 | 25-30 | 30-35 |
|---|---|---|---|---|---|---|
| Frequency | 2 | k | 28 | 54 | k+1 | 5 |
A bag contains 19 unbiased coins and one coin with head on both sides. One coin drawn
at random is tossed and head turns up. If the probability that the drawn coin was unbiased,
is $\frac{m}{n}$, gcd (m, n) = 1, then $n^2 - m^2$ is equal to :
Given three indentical bags each containing 10 balls, whose colours are as follows :
| Red | Blue | Green | |
| Bag I | 3 | 2 | 5 |
| Bag II | 4 | 3 | 3 |
| Bag III | 5 | 1 | 4 |
A person chooses a bag at random and takes out a ball. If the ball is Red, the probability that it is from bag I is $p$ and if the ball is Green, the probability that it is from bag III is $q$, then the value of $\left( \frac{1}{p} + \frac{1}{q} \right)$ is:
| Class | 5-10 | 10-15 | 15-20 | 20-25 | 25-30 | 30-35 |
|---|---|---|---|---|---|---|
| Frequency | 2 | k | 28 | 54 | k+1 | 5 |