A biased coin is tossed twice. If the probability of obtaining a Head is 3/5. a) find the probability distribution of the number of Heads in 2 tosses. b) show that the sum of the probabilities is 1.

Question

anonymous · Answer

On the first toss, you can get H (3/5) or T (2/5).

On the second toss, you can get the same probabilities: H (3/5) or T(2/5).

Put together, you can get HH (3/5 x 3/5 = 9/25), HT (3/5 x 2/5 = 6/25), TH (2/5 x 3/5 = 6/25), or TT (2/5 x 2/5 = 4/25). (The HT and TH events are the same so I'll consider them at the same time.)

Here's a histogram for the frequency distribution:
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anonymous · Answer

If you're concerned with the distribution of heads, then you can consider getting heads to be a success with probability 3/5, and tails a failure with probability 2/5.

Does this success/failure ring any bells? What kind of probability distribution describes this?

anonymous · Answer

i'm not sure

anonymous · Answer

For part B, all you have to do is add the probabilities. (4+9+12)/25 = 25/25 = 1.

anonymous · Answer

ohh thank you :)

anonymous · Answer

What about the binomial distribution? Does that sound familiar?

anonymous · Answer

yes

anonymous · Answer

Okay well you would then use that to model a single coin flip's outcome.

However, a single coin flip has two outcomes, whereas a double coin flip would have three (or four if you consider HT and TH separately), so this might be a multinomial distribution...

Anyway, the name of the probability distribution isn't key here. All you need to do is be able to write a function that models that histogram. Think piecewise function.