91影视

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:

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