Question

A coin is tossed three times. (a) What is the probability that the coin will land tails at least twice? .33 Incorrect: Your answer is incorrect. (b) What is the probability that the coin will land tails on the second toss, given that tails were thrown on the first toss? (c) What is the probability that the coin will land tails on the third toss, given that heads were thrown on the first toss?

Answer 1

a) 0.5

b)0.25

c)0.25

Explanation:

Let H represent the head's outcome

let T represents the outcome of tails.

If the coin is tossed 3times The total number of outcomes are;

HHH, HHT, HTH, HTT, THH, THT, TTH, TTT

Therefore there are total outcomes of 8.

CHECK THE ATTACHMENT FOR DETAILED EXPLANATION

Answer 2

a) P(at least 2 tails) = 1/2

b) P(E₂ | E₁) = 1/2

c) P(E₂ | E₁) = 1/2

Explanation:

We are given that a coin is tossed three times which means there are

2³ = 8 possible outcomes

The sample space is

1. H-H-H

2. T-H-H

3. H-T-H

4. H-H-T

5. T-H-T

6. T-T-H

7. H-T-T

8. T-T-T

(a) What is the probability that the coin will land tails at least twice?

at least twice means 2 or more than 2 tails.

There are exactly 4 number of outcomes where we have at least 2 tails.

P(at least 2 tails) = 4/8

P(at least 2 tails) = 1/2

(b) What is the probability that the coin will land tails on the second toss, given that tails were thrown on the first toss?

The possible outcomes when tail is thrown on first toss are

E₁ = T-H-H, T-H-T, T-T-H, T-T-T (4 outcomes)

The possible outcomes when tail will land on second toss are

E₂ = H-T-H, H-T-T, T-T-H, T-T-T (4 outcomes)

The intersection of these outcomes are

(E₁ ∩ E₂) = T-T-H, T-T-T (2 outcomes)

Therefore, the probability that the coin will land tails on the second toss, given that tails were thrown on the first toss is

P(E₂ | E₁) = (E₁ ∩ E₂) /E₁

P(E₂ | E₁) = 2/4

P(E₂ | E₁) = 1/2

(c) What is the probability that the coin will land tails on the third toss, given that heads were thrown on the first toss?

The possible outcomes when head is thrown on first toss are

E₁ = H-H-H, H-H-T, H-T-H, H-T-T (4 outcomes)

The possible outcomes when tail will land on third toss are

E₂ = H-H-T, H-T-T, T-H-T, T-T-T (4 outcomes)

The intersection of these outcomes are

(E₁ ∩ E₂) = H-H-T, H-T-T (2 outcomes)

Therefore, the probability that the coin will land tails on the third toss, given that heads were thrown on the first toss is

P(E₂ | E₁) = (E₁ ∩ E₂) /E₁

P(E₂ | E₁) = 2/4

P(E₂ | E₁) = 1/2