
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
162.6k+ views
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.
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.
Recently Updated Pages
Fluid Pressure - Important Concepts and Tips for JEE

JEE Main 2023 (February 1st Shift 2) Physics Question Paper with Answer Key

Impulse Momentum Theorem Important Concepts and Tips for JEE

Graphical Methods of Vector Addition - Important Concepts for JEE

JEE Main 2022 (July 29th Shift 1) Chemistry Question Paper with Answer Key

JEE Main 2023 (February 1st Shift 1) Physics Question Paper with Answer Key

Trending doubts
Degree of Dissociation and Its Formula With Solved Example for JEE

IIIT JEE Main Cutoff 2024

IIT Full Form

JEE Main Reservation Criteria 2025: SC, ST, EWS, and PwD Candidates

JEE Main Cut-Off for NIT Kurukshetra: All Important Details

JEE Main Cut-Off for VNIT Nagpur 2025: Check All Rounds Cutoff Ranks

Other Pages
NEET 2025: All Major Changes in Application Process, Pattern and More

Verb Forms Guide: V1, V2, V3, V4, V5 Explained

NEET Total Marks 2025: Important Information and Key Updates

1 Billion in Rupees - Conversion, Solved Examples and FAQs

NEET 2025 Syllabus PDF by NTA (Released)

Important Days In June: What Do You Need To Know
