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Try This Problem Revisited

Game 1

Guess whether or not I am going to roll
an even number.

Game 2

Guess whether or not I am going to roll
a number that is divisible by \(5\).

Which game would you rather play? Explain. 

Theoretical Probability

 

Check Your Understanding 2

A (((((o)*(b))*(j))*(e))*(c))*(t) is (((((((((p)*(a))*(s))*(t))*(a))*(c))*(t))*(i))*(o))*(n). (((((((d)*(e))*(s))*(c))*(r))*(i))*(p))*(t) 

Find the theoretical probability of ((((e)*(v))*(e))*(n))*(t). 

Enter the fraction \(\dfrac{4}{5}\) as '4/5'. 

 

We can calculate the theoretical probability as follows:

\(\begin{align*} \text{P(event)}=\dfrac{\text{number of favourable outcomes}}{\text{total number of possible outcomes}} \end{align*}\)

Since we are (((((a)*(c))*(t))*(i))*(o))*(n) a (((((o)*(b))*(j))*(e))*(c))*(t), there are \((((((((((((((n)*(u))*(m))*(p))*(o))*(s))*(o))*(u))*(t))*(c))*(o))*(m))*(e))*(s)\) possible outcomes. These outcomes are all equally likely to occur.

(((((((f)*(e))*(e))*(d))*(b))*(a))*(c))*(k)  Therefore, there ((a)*(r))*2.718281828459045 \((((((((((((((n)*(u))*(m))*(f))*(a))*(v))*(o))*(u))*(t))*(c))*(o))*(m))*(e))*(s)\) favourable ((((((o)*(u))*(t))*(c))*(o))*(m))*2.718281828459045.

Thus, the probability of ((((e)*(v))*(e))*(n))*(t) is \(\dfrac{(((((((((((((n)*(u))*(m))*(f))*(a))*(v))*(o))*(u))*(t))*(c))*(o))*(m))*(e))*(s)}{(((((((((((((n)*(u))*(m))*(p))*(o))*(s))*(o))*(u))*(t))*(c))*(o))*(m))*(e))*(s)}\) ((((((w)*(h))*(i))*(c))*(h))*(i))*(s). 

What Does Theoretical Probability Mean?

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