Coin probability

Question

A loaded coin, with P(H)= 3/5 and P(T)=2/5 on each single toss, is tossed 6 times. Let E be the event that exactly 3 heads appeared in 6 tosses and F be the event that at least 2 heads appeared in 6 tosses. Determine the following probabilities:

A) P(E)=

B) P(F)=

C) P(EnF)=

D) P(E/F)=

Answer 1

We need to determine the number of possible outcomes after tossing 6 coins.  I use (a + b)ⁿ with a = 1 and b = 1.  This will give you 2ⁿ.  Let n = 6.  Then 2⁶ = 64 possible outcomes.

 

Next, we will use the formula for combinations:  n! / x!(n - x)! where x is the number of heads.

 

Let p = the probability of getting a head on a single toss and let q = the probability of getting a tail on a single toss.

then P(H) = 3/5 and P(T) = 2/5.

 

This is the probability used in the binomial distribution.

 

[n! / x!(n - x)!]*p^xq^n-x

 

A)  P(E) = [6!/(3!)(3!)]*(3/5)³(2/5)³ = 0.276

 

B)  P(F) = [6!/(2!)(4!)]*(3/5)²(2/5)⁴ + [6!/(3!)(3!)]*(3/5)³(2/5)³ + [6!/(4!)(2!)]*(3/5)⁴(2/5)² + [6!/(5!)(1!)]*(3/5)⁵(2/5)¹ + [6!/(6!)(0!)]*(3/5)⁶(2/5)⁰

 

= 0.138 + 0.276 + 0.311 + 0.187 + 0.0467 = 0.9587

 

C)  P(E and F) = P(E)*P(F) = (0.276)(0.9587) = 0.265.

This is the probability of getting at least 2 heads that appeared in 6 tosses and getting exactly 3 heads that appeared in 6 tosses. 

 

D)  P(E|F) = P(E and F)/P(F) = 0.265/0.9587 = 0.276.

This is the probability of getting exactly 3 heads that appeared in 6 tosses given that you will get at least 2 heads that will appear in 6 tosses.