Question

Question:\(P({\rm{ tails }}) = \frac{1}{4},P({\rm{ heads }}) = \frac{3}{4}\)

a) What conditions should be met by the probabilities assigned to the outcomes from a finite sample space?

b) What probabilities should be assigned to the outcome of heads and the outcome of tails if heads come up three times as often as tails?

Short answer

Answer

(a) The resultant answer isthat all probabilities are between 0 and 1 including.The sum of the probabilities of all possible outcomes needs to be equal to 1.

(b) The resultant answer is \(P({\rm{ tails }}) = \frac{1}{4}\), \(P({\rm{ heads }}) = \frac{3}{4}\)

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Step-by-step solution

Given data

The given datais finite sample space, heads and tails.

Concept of Probability 

Finding the probability of an event occurring can be thought of as probability.

Write the properties of finite sample space

(a)

The probabilities in a finite sample space need to have the following properties:

- All probabilities are between 0 and 1 including.

- The sum of the probabilities of all possible outcomes needs to be equal to 1.

Find the probability of first 50 positive integer

(b)

Heads comes up three times as often as tails.

This then implies that the probability of heads to be three times the probability of tails:

\(P({\rm{ heads }}) = 3P({\rm{ tails }}).\)

The sum of the probabilities of all possible outcomes needs to be equal to 1

\(P({\rm{ heads }}) + P({\rm{ tails }}) = 1\)

Since \(P({\rm{ heads }}) = 3P({\rm{ tails }}).\)

\(3P({\rm{ tails }}) + P({\rm{ tails }}) = 1\)

Combine like terms:\(4P({\rm{ tails }}) = 1\)

Divide each side of the previous equation by 4:

\(P({\rm{ tails }}) = \frac{1}{4}\)

Finally, we obtain the probability of heads by evaluating the equation

\(P({\rm{ heads }}) = 3P({\rm{ tails }}).\)

\(\begin{aligned}{}P({\rm{ heads }}) &= 3P({\rm{ tails }})\\ &= 3 \cdot \frac{1}{4}\\&= \frac{3}{4}\end{aligned}\)