Consider 4 flips of an unfair coin where probability of heads is 0.40. Consider all tosses independent of each other. (a) Compare the probability of 4 heads in 4 tosses and 4 tails in 4 tosses. (b) Determine the probability of H T H T in 4 tosses. (c) Determine the number of outcomes that give us 2 heads in 4 tosses in any order. What is the probability of 2 heads in 4 tosses (use (b) and the number of outcomes)

a) 0.0256 = 2.56% probability of 4 heads.

0.1296 = 12.96% probability of 4 tails.

b) The probability is 0.0576 = 5.76%.

c) 6 outcomes given 2 heads in 4 tosses. 0.3456 = 34.56% probability of 2 heads in 4 tosses

Step-by-step explanation:

For each throw, we have these following probabilities:

0.4 = 40% probability of heads.

0.6 = 60% probability of tails.

(a) Compare the probability of 4 heads in 4 tosses and 4 tails in 4 tosses.

4 heads:

4 throws, each with 0.4 probability of heads. So

0.0256 = 2.56% probability of 4 heads.

4 tails:

4 throws, each with 0.6 probability of tails. So

0.1296 = 12.96% probability of 4 tails.

(b) Determine the probability of H T H T in 4 tosses.

H = 0.4, T = 0.6. So

The probability is 0.0576 = 5.76%.

(c) Determine the number of outcomes that give us 2 heads in 4 tosses in any order. What is the probability of 2 heads in 4 tosses (use (b) and the number of outcomes)

Number of outcomes:

The possible outcomes are:

H - H - T - T

H - T - H - T

H - T - T - H

T - H - H - T

T - H - T - H

T - T - H - H

6 outcomes given 2 heads in 4 tosses.

Each with probability 0.0576, as found in b)

6*0.0576 = 0.3456

0.3456 = 34.56% probability of 2 heads in 4 tosses

sup

step-by-step explanation:

depending on others is critical to survival. my source is apex!

step-by-step explanation: