Question

A coin is tossed n times. If the probability of getting head at least once is greater than 0.8, then the least value of n is
A. 2
B. 3
C. 5
D. 4

Answer 1

Hint: Find the probability of getting heads and getting tails in one toss. Then find the probability of getting heads in the first toss, then the probability of getting heads for the first time in the second toss, and so on. Add all these probabilities to find the probability of getting heads at least once.

Formula Used: \[{\text{Probability of an event = }}\dfrac{{{\text{Number of favourable outcomes}}}}{{{\text{Total number of outcomes}}}}\]

Complete step by step Solution:
Let the event of getting heads be H and the event of getting tails be T.
If a coin is tossed, the probability of getting heads, $P\left( H \right) = \dfrac{1}{2}$ and the probability of getting tails, $P\left( T \right) = \dfrac{1}{2}$.
Let the probability of getting heads in the first toss be \[P\left( {{H_1}} \right)\], the probability of getting heads for the first time in the second toss be $P\left( {{H_2}} \right)$ and the probability of getting heads for the first time in the ${n^{th}}$ toss be $P\left( {{H_n}} \right)$.
Probability of getting head at least once is the sum of \[P\left( {{H_1}} \right)\], $P\left( {{H_2}} \right)$…. $P\left( {{H_n}} \right)$.
Probability = \[P\left( {{H_1}} \right) + P\left( {{H_2}} \right) + .... + P\left( {{H_n}} \right) = P\left( H \right) + P\left( T \right)P\left( H \right) + ... + P{\left( T \right)^{n - 1}}P\left( H \right)\]
Probability = $\dfrac{1}{2} + \dfrac{1}{4} + \dfrac{1}{8} + ......$
Since the probability is greater than 0.8,
$\dfrac{1}{2} + \dfrac{1}{4} + \dfrac{1}{8} + ...... \geqslant 0.8$
Using sum of GP we get:
$\dfrac{{\dfrac{1}{2}\left( {1 - {{\left( {\dfrac{1}{2}} \right)}^n}} \right)}}{{1 - \dfrac{1}{2}}} \geqslant \dfrac{4}{5}$
$1 - {\left( {\dfrac{1}{2}} \right)^n} \geqslant \dfrac{4}{5}$
$1 - \dfrac{4}{5} \geqslant {\left( {\dfrac{1}{2}} \right)^n}$
$\left( {\dfrac{1}{{{2^n}}}} \right) \leqslant \dfrac{1}{5}$
Solving further, we get ${2^n} \geqslant 5$.
If n is 2 then the above expression would be false.

Therefore, the correct option is (B).

Note: Alternate way – The probability of getting heads at least once is the complement of getting all tails. After n tosses the probability of getting all tails is $\dfrac{1}{{{2^n}}}$. Therefore, the probability of getting at least one head is $1 - \dfrac{1}{{{2^n}}}$. Solve the inequality, $1 - {\left( {\dfrac{1}{2}} \right)^n} \geqslant \dfrac{4}{5}$ to get n.