Let's examine ONE case in which we get exactly 3 heads: HHHTT

P(HHHTT) = (1/2)(1/2)(1/2)(1/2)(1/2) = 1/32

This, of course, is just ONE possible way to get exactly 3 heads.

Another possible outcome is HHTTH

Here, P(HHTTH) = (1/2)(1/2)(1/2)(1/2)(1/2) = 1/32

As you might guess, each possible outcome will have the same probability (1/32). So, the question becomes "In how many different ways can we get exactly 3 heads and 2 tails?"

In other words, in how many different ways can we arrange the letters HHHTT?

Well, we can apply the MISSISSIPPI rule (see video below) to see that the number of arrangements = 5!/(3!)(2!) = 10

So P(exactly 3 heads) = (1/32)(10) = 10/32 = 5/16

RELATED VIDEO

Would u like to explain the below problem as like as u solve the above?
A fair coin is tossed 5 times. What is the probability of getting at least three heads on consecutive tosses?
Specifically, I ask u coz I like ur every explanation so much.

Hi Brent,

I am stuck with the approach


Ques: A fair coin is tossed 5 times. What is the probability of getting at least three heads on consecutive tosses.

My take::

Probability of Head or Tails for each coin flip -> (1/2) Multiply by number of coin flips -> (1/2)(1/2)(1/2)(1/2)(1/2) = 1/32

Now we need to find the chance of getting three heads consecutively

So we have 8 scenarios where at LEAST three heads will occur consecutively -> HHHTT, HHHTH, HHHHT, HHHHH, THHHT, THHHH, TTHHH, HTHHH

probability= 8/32 = 1/4


I tried ur approach and got,

P(exactly 3 heads) = 10/32
P(exactly 4 heads) = 5/32
P(exactly 5 heads) = 1/32

Probability = 16/32 = 1/2

am I getting wrong somewhere

So we have 8 scenarios where at LEAST three heads will occur consecutively -> HHHTT, HHHTH, HHHHT, HHHHH, THHHT, THHHH, TTHHH, HTHHH
probability= 8/32 = 1/4