Question

If a coin is tossed \(10\) times, the probability of getting head at least two times but at most five times is

• A. \(\dfrac{14}{10^2}\)
• B. \(\dfrac{627}{2^{10}}\)
• C. \(\dfrac{2^{10}-627}{2^{10}}\)
• D. \(\dfrac{{}^{10}C_2+{}^{10}C_3+{}^{10}C_4+{}^{10}C_5}{2^{10}}\)

Hint:

For a fair coin tossed \(n\) times, \[ P(X=r)=\frac{{}^{n}C_r}{2^n} \] where \(X\) is the number of heads. For probabilities involving a range of values, add the corresponding binomial probabilities.

Answer 1

The Correct Option is B

Step 1: Model the tosses.
Each toss of a fair coin is independent with head-probability $\tfrac12$. The number of heads $X$ in $10$ tosses follows a binomial pattern, and $P(X=r)=\dfrac{{}^{10}C_r}{2^{10}}$.

Step 2: Translate the words.
"At least two but at most five heads" means $X=2,3,4$ or $5$.

Step 3: Write the probability as a sum.
$P(2\le X\le5)=\dfrac{{}^{10}C_2+{}^{10}C_3+{}^{10}C_4+{}^{10}C_5}{2^{10}}$.

Step 4: Compute the coefficients.
${}^{10}C_2=45$, ${}^{10}C_3=120$, ${}^{10}C_4=210$, ${}^{10}C_5=252$.

Step 5: Add them.
$45+120+210+252=627$.

Step 6: State the probability.
So the required probability is $\dfrac{627}{2^{10}}$. \[ \boxed{\dfrac{627}{2^{10}}} \]