The tree diagram on page 7 of Chapter 7 shows a sequence of 4 Bernoulli trials,

Question

The tree diagram on page 7 of Chapter 7 shows a sequence of 4 Bernoulli trials, with a constant probability of landing on the carpet of 0.9. Let us focus on the sequences that begin with a carpet landing and the next 3 landings include at least 1 slime landing. What is that probability? Hint: If necessary, sum the relevant probabilities you calculated from the tree diagram.?

A. 0.0729

B. 0.0252

C. 0.8748

D. 0.0243

E. 0.0009

F. 0.3439

Tree diagram of probabilities of intersections of 4 independent slide trials with a fixed probability of 0.9 of a carpet landing Intersection Probability 0.6561 09 carpet Slime Carpet Slime Carpet Slime Carpet Slime C1, C2, C3, C4 0.9Carpet 0.9 Carpet Slime Carpet Carpet 0.9 0.1 Slime Slime Carpet Slime Carpet Slime Carpet Slime Carpet Slime Carpet Carpet 0.1 0.9 Slime Slime Slime Carpet 0.1 Slime

Explanation / Answer

Since it given that first landing is carpet so it must be upper half.

Following is the sequence:

P(carpet, carpet,carpet,slime) = 0.9*0.9*0.9*0.1 = 0.0729

P(carpet, carpet,slime, carpet) = 0.9*0.9*0.1*0.9 = 0.0729

P(carpet, carpet,slime, slime) = 0.9*0.9*0.1*0.1 = 0.0081

P(carpet, slime, carpet, carpet) = 0.9*0.1*0.9*0.9 = 0.0729

P(carpet, slime, carpet, slime) = 0.9*0.1*0.9*0.1 = 0.0081

P(carpet, slime, slime, carpet) = 0.9*0.1*0.1*0.9 = 0.0081

P(carpet, slime, slime, slime) = 0.9*0.1*0.1*0.1 = 0.0009

Adding all above give the required probability. So required probability is 0.2439.

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