A loaded coin is tossed 5 times, find the probability distribution for the number of heads, if heads are thrice likely to occur than tails. Let x be the number of heads that will turn up. Compute the probability associated to each value of X using the equation:
x=0,1,2,3,4,5
P(X=x)= (5/4)(3/4)x(1/4)5-x note x and 5-x outside () are superscripts there..
To compute the probability distribution for the number of heads when a loaded coin is tossed 5 times, we can use the equation provided:
P(X=x) = (5/4) * (3/4)^x * (1/4)^(5-x)
Where:
- P(X=x) is the probability of getting x heads,
- (5/4) is the overall probability scale factor,
- (3/4)^x represents the likelihood of getting x heads (since heads are 3 times more likely than tails),
- (1/4)^(5-x) represents the likelihood of getting 5-x tails.
Now, let's calculate the probability for each value of x:
For x = 0:
P(X=0) = (5/4) * (3/4)^0 * (1/4)^(5-0)
= (5/4) * 1 * 1
= 5/4
= 1.25
For x = 1:
P(X=1) = (5/4) * (3/4)^1 * (1/4)^(5-1)
= (5/4) * (3/4) * (1/4)^4
= (5/4) * (3/4) * 1/256
= 15/1024
≈ 0.0146
For x = 2:
P(X=2) = (5/4) * (3/4)^2 * (1/4)^(5-2)
= (5/4) * (3/4)^2 * (1/4)^3
= (5/4) * (9/16) * 1/64
= 45/1024
≈ 0.0439
For x = 3:
P(X=3) = (5/4) * (3/4)^3 * (1/4)^(5-3)
= (5/4) * (3/4)^3 * (1/4)^2
= (5/4) * (27/64) * 1/16
= 135/4096
≈ 0.0329
For x = 4:
P(X=4) = (5/4) * (3/4)^4 * (1/4)^(5-4)
= (5/4) * (3/4)^4 * 1/4
= (5/4) * (81/256) * 1/4
= 405/4096
≈ 0.0987
For x = 5:
P(X=5) = (5/4) * (3/4)^5 * (1/4)^(5-5)
= (5/4) * (3/4)^5 * 1
= (5/4) * (243/1024)
= 1215/4096
≈ 0.2969
So, the probability distribution for the number of heads (X) when a loaded coin is tossed 5 times is as follows:
P(X=0) ≈ 1.25
P(X=1) ≈ 0.0146
P(X=2) ≈ 0.0439
P(X=3) ≈ 0.0329
P(X=4) ≈ 0.0987
P(X=5) ≈ 0.2969