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2. Toss 4 times Imagine tossing a fair coin 4 times . (a ) Give a probability model for this chance process . (b) Define event B as getting exactly three tails . Find P(B)

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Answer:

If a coin is tossed 4 times; number of possible outcomes = (2^4) = 16. 3 heads and 1 tail can occur from 4 tosses in (4C3)*(1C1) = 4*1 = 4 ways. So, the probability = (4 / 16) = (1 / 4) = 0.25.  

i didnt really understand the question but hope this helps

According to the information given, we have that:

  • a) A binomial model with [tex]p = 0.5[/tex] and [tex]n = 4[/tex] should be used.
  • b) P(B) = 0.25.

For each coin, there are only two possible outcomes, either it is heads, or it is tails. The result of the flip of a coin is independent of any other flip, hence, the binomial distribution is used to solve this question.

Binomial probability distribution

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]

The parameters are:

  • x is the number of successes.
  • n is the number of trials.
  • p is the probability of a success on a single trial.

Item a:

  • The coin is tossed 4 times, hence [tex]n = 4[/tex].
  • It is a fair coin, hence equally as likely to be heads and tails, which means that [tex]p = 0.5[/tex].

Hence, a binomial model with [tex]p = 0.5[/tex] and [tex]n = 4[/tex] should be used.

Item b:

P(B) is P(X = 3), hence:

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

[tex]P(X = 3) = C_{4,3}.(0.5)^{3}.(0.5)^{1} = 0.25[/tex]

Hence, P(B) = 0.25.

To learn more about the binomial distribution, you can take a look at https://brainly.com/question/24863377