JEE MAIN - Mathematics (2019 - 8th April Evening Slot - No. 16)
The minimum number of times one has to toss a
fair coin so that the probability of observing at least
one head is at least 90% is :
2
3
4
5
Explanation
Probablity of getting head P(H) = $${1 \over 2}$$
Probablity of getting tail P(T) = 1 - $${1 \over 2}$$ = $${1 \over 2}$$
Probability of observing at least one head out of n tosses
= 1 - Probability of observing no head occurs out of n tosses
= 1 - $${\left( {{1 \over 2}} \right)^n}$$
According to the question,
1 - $${\left( {{1 \over 2}} \right)^n}$$ $$ \ge $$ $${{90} \over {100}}$$
$$ \Rightarrow $$ $${\left( {{1 \over 2}} \right)^n} \le {1 \over {10}}$$
$$ \Rightarrow $$ $${2^n} \ge 10$$
As 23 = 8
24 = 16
$$ \therefore $$ Minimum value of n = 4
Probablity of getting tail P(T) = 1 - $${1 \over 2}$$ = $${1 \over 2}$$
Probability of observing at least one head out of n tosses
= 1 - Probability of observing no head occurs out of n tosses
= 1 - $${\left( {{1 \over 2}} \right)^n}$$
According to the question,
1 - $${\left( {{1 \over 2}} \right)^n}$$ $$ \ge $$ $${{90} \over {100}}$$
$$ \Rightarrow $$ $${\left( {{1 \over 2}} \right)^n} \le {1 \over {10}}$$
$$ \Rightarrow $$ $${2^n} \ge 10$$
As 23 = 8
24 = 16
$$ \therefore $$ Minimum value of n = 4
Comments (0)
