Question

Q2. A man wins in a gambling game if he gets two heads in in five flips
of a biased coin. The probability of getting a head with the coin is 0.7.
1. Find the probability the man will win. Should he play this game?
2. What is the probability of winning if he wins by getting at least four
heads in five flips? Should he play this new game?
​

Answer 1

Answer:

The probability the man will win will be 13.23%. And the probability of winning if he wins by getting at least four heads in five flips will be 36.01%.

Step-by-step explanation:

Step 1:  Binomial distributions consist of n independent Bernoulli trials.

Bernoulli trials are those trials that end up randomly either on success (with probability p) or on failures( with probability 1- p = q (say))

P(X = x) = ⁿCₓ pˣ (1 - p)⁽ⁿ⁻ˣ⁾

Step 2: A man wins in a gambling game if he gets two heads in five flips of a biased coin. the probability of getting a head with the coin is 0.7.

Then we have

p = 0.7

n = 5

Step 3:Then the probability the man will win will be

P(X = 2) = ⁵C₂ (0.7)² (1 - 0.7)⁽⁵⁻²⁾

P(X = 2) = 10 x 0.49 x 0.027

P(X = 2) = 0.1323

P(X = 2) = 13.23%

Then the probability of winning if he wins by getting at least four heads in five flips will be

P(X = 4) = ⁵C₄ (0.7)⁴ (1 - 0.7)⁽⁵⁻⁴⁾

P(X = 4) = 5 x 0.2401 x 0.3

P(X = 4) = 0.3601

P(X = 4) = 36.01%

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Answer 2

Answer:

Step-by-step explanation: