Q1. A fair coin is tossed 2019 times. Probability of getting 1010 heads is-

Q2. From the first 3n(n greater than or equal to 2) natural numbers two numbers are picked up at random without replacement. Find probability that the absolute value of difference between the numbers is less than n -

Q3. The minimum number of time a fair coin needs to be tossed , so that the probability of getting atleast two heads is atleast 0.96 is-

# Doubt from probability

- 1/2 ?

Yes , please share your method

3rd question 8?

See, upto 2018 tosses for sure I will get 1009 Heads and 1009 Tails , as it is a fair die. Now the 2019th toss will give me either head or tail with probability 1/2 . So if I get Head then total head becomes 1010 . Hence probability is 1/2.

2 Likes

Yes bro

Not correct

What's the mistake?

yes bro sending

A fair coin can also give you 2018 tails or heads

It is a binomial distribution which will give answer

$$\binom{2018}{1010}/2^{2018}$$