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

Concept: For \(n\) tosses of a fair coin, \[ P(X=r)=\frac{{}^{n}C_r}{2^n} \] where \(X\) denotes the number of heads obtained.

Step 1: Express the required probability. We need \[ P(2\le X\le 5) \] Hence, \[\begin{aligned} P(2\le X\le 5) = \frac{ {}^{10}C_2+ {}^{10}C_3+ {}^{10}C_4+ {}^{10}C_5 }{2^{10}} \end{aligned}\]

Step 2: Evaluate the binomial coefficients. \[\begin{aligned} {}^{10}C_2 &=45\\ {}^{10}C_3 &=120\\ {}^{10}C_4 &=210\\ {}^{10}C_5 &=252 \end{aligned}\] Therefore, \[\begin{aligned} 45+120+210+252 = 627 \end{aligned}\]

Step 3: Find the probability. \[\begin{aligned} P(2\le X\le 5) &= \frac{627}{2^{10}} \end{aligned}\] \[\begin{aligned} \boxed{\frac{627}{2^{10}}} \end{aligned}\] Hence, option \(\mathbf{(B)}\) is correct.