Question

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:

(a) Write the event “plays with at most two toys” in terms of$X$. What is the probability of this event?

(b) Describe the event$X>3$ in words. What is its probability? What is the probability that $X\ge 3$?

Short answer

(a)The probability is$0.49$

(b)$P(X>3)=0.28$

$P(X\ge 3)=0.51$

Step-by-step solution

Given information(part a)

Given in the question that, 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:

We need to find the probability of the event “plays with at most two toys” .

Explanation(part a)

The probability distribution is:

Value of $X$	$0$	$1$	$2$	$3$	$4$	$5$
Probability	$0.03$	$0.16$	$0.30$	role="math" localid="1649686470559" $0.23$	$0.17$	$0.11$

Consider, $\mathrm{X}$be the random variable that shows the number of toys boys play. The event "plays with at most$2$toys" could be written as $X\le 2$.

The probability is computed as:

$P(X\le 2)=P(X=0)+P(X=1)+P(X=2)$

$=0.03+0.16+0.30$

$=0.49$

The probability is $0.49$.

Given information(part b)

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:

We need to find the probability of $X>3$and $X\ge 3$.

Explanation(part b)

The event $X>3$means that boy plays with more than 3 toys. The probabilities could be calculated as:

$P(X\ge 3)=P(X=3)+P(X=4)+P(X=5)$

$=0.23+0.17+0.11$

$=0.51$

Now,

$P(X>3)=P(X=4)+P(X=5)$

$=0.17+0.11$

$=0.28$

Thus, the probabilities are $0.51$ and $0.28$ respectively.