Practicing Success
If a fair coin is tossed 10 times, then the probability of obtaining at least one head is : |
$\frac{1}{1024}$ $\frac{17}{1024}$ $\frac{1023}{1024}$ $\frac{23}{1024}$ |
$\frac{1023}{1024}$ |
P(atleast one head) = P(one head) + P(two heads) + P(three heads) + ... + P(ten heads) By Bernoulli's probability distribution eg: for obtaining (1) head in 10 toses → no of possible ways $={ }^{10} C_1$ for obtaining (2) Heads → possible ways $={ }^{10} C_2$ probability for each case = $\frac{1}{2} \times \frac{1}{2} ... \frac{1}{2}$ (10 times) $=\frac{1}{2^{10}}$ so, total probability of atleast 1 head $=\frac{10}{{ }^{10} C_{10}}+\frac{10 C_2}{2^{10}}+...\frac{{ }^{10} C_{10}}{2^{10}}$ P(atleast 1 head) = $\frac{{ }^{10} c_1+{ }^{10} c_2+...{ }^{10} c_{10}}{2^{10}}$ we know that $2^n=(1+1)^n={ }^n c_0+{ }^n c_1+{ }^n c_2...{ }^n c_n$ so $2^n=1+{ }^n c_1+{ }^n c_2 ...{ }^n c_n$ $\Rightarrow\left(2^n-1\right)={ }^n c_1+{ }^n c_2 ...{ }^n c_n$ so P(atleast 1 head) = $\frac{2^{10}-1}{2^{10}}=\frac{1024-1}{1024}=\frac{1023}{1024}$ |