I have listed the question and the professors answer below. Can you please show

Question

I have listed the question and the professors answer below. Can you please show me how to solve this problem? Thank you so much

QUESTION: In the game of Heads or Tails, Betty and Bob toss a coin four times. Betty wins a dollar from Bob for each head and pays Bob a dollar for each tail - that is, she wins or loses the difference between the number of heads and the number of tails. For example, if there are one head and 3 tails, Betty loses $2. You can check that Betty's possible outcomes are {-4,-2,0,2,4}. Assign probabilities to these outcomes by playing the game 20 times and using the proportions of the outcomes as estimates of the probability. If possible, combine your trials with those of other students to obtain long-run proportions that are closer to the probabilities. (M&M 4.7)

ANSWER: Thetheoretical probabilitiesare(inorder): 1/16,4/16,6/16,4/16,1/16.

Explanation / Answer

The experimental probability that you will get may vary as per the outcome of the toss

But the theoretical or long run probability will be the same. Its calculated as shown below

To lose $4, all 4 tosses must result in tails

P(tails) = 1/2

So, P(4 tails) = 1/2 x 1/2 x 1/2 x 1/2 = 1/16

To lose $2, there shall be 3 tails and 1 head

Number of cases possible = 4C1 = 4!/(3!x1!) = 4

The outcomes are TTTH, TTHT, THTT and HTTT

All 4 outcomes have a probability of 1/16

So, sum of these 4 probabilities = 4x1/16 = 4/16

For net outcome to be $0, there shall be 2 heads and 2 tails

Number of cases possible = 4C2 = 4!/(2!x2!)

= 6

So, probability = 6x1/16 = 6/16

For winning $2, the probability is 4C3 x 1/16

= 4x1/16

= 4/16

P(winning $4) = P(all heads) = 1/2 x 1/2 x 1/2 x 1/2 = 1/16

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