Kids and toys In an experiment on the behavior of young children, each subject is placed in an area with five toys. Past experiments have shown that the probability distribution of the number X of toys played with by a randomly selected subject is as follows:

Part (a). Write the event “child plays with 5 toys” in terms of X. Then find its probability.

Part (b). What’s the probability that a randomly selected subject plays with at most 3 toys?

Let us refer to the missing probability as x.

Let X represent the number of toys the child has. X = 5 represents the event "child plays with 5 toys."

If all probabilities are between 0 and 1, the probability distribution is valid (including).

The total of the probabilities of each outcome equals one.

The sum of all probabilities in the preceding table must then equal 1.

$$\begin{gathered} 0.03+0.16+0.30+0.23+0.17+x=1 \\ 0.89+x=1 \end{gathered}$$

take 0.89 off each side

$$\begin{gathered} x=1-0.89 \\ x=0.11 \end{gathered}$$

As a result, the probability of a child playing with 5 toys is 0.11.

$P\left(x=5\right)=0.11$

If two events cannot occur at the same time, they are disjoint or mutually exclusive.

Addition rule for events that are disjoint or mutually exclusive:

$P\left(AUB\right)=P\left(A\right)+P\left(B\right)$

Let us refer to the missing probability as x.

In the above table, we can calculate the probability of 0, 1, 2, and 3 toys:

$$\begin{gathered} P(X=0)=0.03 \\ P(X=1)=0.16 \\ P(X=2)=0.30 \\ P(X=3)=0.23 \end{gathered}$$

The events are mutually exclusive because a child cannot play with two different amounts of toys.

For mutually exclusive events, apply the addition rule:

$$\begin{gathered} P(X\le 3)=P(X=0)+P(X=1)+P(X=2)+P(X=3) \\ P(X\le 3)=0.03+0.16+0.30+0.23 \\ P(X\le 3)=0.72 \\ P(X\le 3)=72\% \end{gathered}$$