Question

3 coined toised simultaneously find the probelity of getting atleast two heads​

Answer 1

Answer:

Possible outcomes of tossing three coins are:

(HHH), (HHT), (HTH), (THH), (TTT), (TTH), (THT),(HTT)

here H and T are denoted for Head and Tail.

Total outcomes = 8

no. of outcomes with at least two heads = 4

∴ required probability =  

8 /4 = 2 /1

​  

Answer 2

Here is the simplest demonstration: If you flip a coin three times, you’re either going to get more heads or more tails. There is no other possibility. Assuming the coin is unbiased, a majority-heads result can be no more or less likely than a majority-tails result. Therefore the answer is 1/2.

You can also get this by looking at the third row of Pascal’s Triangle, which gives you the binomial coefficients:

1 3 3 1

which implies:

Ways of getting 0 heads = 1

Ways of getting 1 head = 3

Ways of getting 2 heads = 3

Wasy of getting 3 heads = 1

So… there are 4 ways of getting two or more heads (3 + 1). Divided this by the total distribution:

(3 + 1 ) / (1 + 3 + 3 + 1) = 4 / 8 = 1/2.

There’s your answer.