Question

The relative frequency table shows the results of a survey in which parents were asked how much time their children spend playing outside and how much time they spend using electronics.
Given that a child spends at least 1 hour per day outside, what is the probability, rounded to the nearest hundredth if necessary, that the child spends less than 1 hour per day on electronics?

Answer 1

Answer: 0.88

Step-by-step explanation:

Let A denotes the number of children spend at least 1 hour per day outside and B denote the number of children spend less than 1 hour per day on electronics .

From the given  relative frequency table , we have

The total number of children spend at least 1 hour per day outside = 16

Probability of a child spends at least 1 hour per day outside is given by :-

$\text{P(A)}=\dfrac{16}{64}$

The total number of children spend less than 1 hour per day on electronics and spend at least 1 hour per day outside = 14

Probability of a child spends less than 1 hour per day on electronics and at least 1 hour per day outside is given by :-

$P(A\cap B)}=\dfrac{14}{64}$

The conditional probability of B , given that A is given by :-

$P(B|A)=\dfrac{P(A\cap B)}{P(A)}\\\\\Rightarrow\ P(B|A)=\dfrac{\frac{14}{64}}{\frac{16}{64}}\\\\\Rightarrow\ P(B|A)=\dfrac{14}{16}=0.875\approx0.88$

Hence, the probability that the child spends less than 1 hour per day on electronics =0.88

Answer 2

total 16 children spend at least 1 hour/day outside
and 14 out of 16 spend less than 1 hour/day using electronics

so 14/16 = 0.875 = 87.5% = 88%(rounded to nearest percent)

answer
88%