Answer :
To find the probability that a child spends less than 1 hour per day on electronics given that the child spends at least 1 hour per day outside, we need to use the concept of conditional probability.
Conditional probability is the likelihood of an event occurring given that another event has already occurred. In this case, we are interested in the probability that a child spends less than 1 hour per day on electronics (Event A), given that the child spends at least 1 hour per day outside (Event B).
The formula for conditional probability [tex]\( P(A|B) \)[/tex] is given by:
[tex]\[ P(A|B) = \frac{P(A \cap B)}{P(B)} \][/tex]
Where:
- [tex]\( P(A \cap B) \)[/tex] is the probability that both Event A and Event B occur.
- [tex]\( P(B) \)[/tex] is the probability that Event B occurs.
From the given table, the total number of children surveyed is 64. Among them, the number of children who spend at least 1 hour per day outside is 16. Therefore, [tex]\( P(B) \)[/tex] is the probability that a child spends at least 1 hour per day outside.
To find [tex]\( P(A \cap B) \)[/tex]:
- We observe that there are 14 children who spend less than 1 hour per day on electronics and also spend at least 1 hour per day outside.
So:
[tex]\[ P(A \cap B) = \frac{\text{Number of children who spend less than 1 hour per day on electronics and spend at least 1 hour per day outside}}{\text{Total number of children}} = \frac{14}{64} \][/tex]
[tex]\[ P(B) = \frac{\text{Number of children who spend at least 1 hour per day outside}}{\text{Total number of children}} = \frac{16}{64} \][/tex]
Note that in conditional probability, instead of the overall total, we consider the total number of children who meet the condition (Event B). Therefore, the actual calculation involves the numbers directly:
[tex]\[ P(A|B) = \frac{14}{16} \][/tex]
By performing the division, we get:
[tex]\[ P(A|B) = \frac{14}{16} = 0.875 \][/tex]
Finally, rounding this probability to the nearest hundredth:
[tex]\[ P(A|B) \approx 0.88 \][/tex]
So, the probability that a child spends less than 1 hour per day on electronics given that the child spends at least 1 hour per day outside is approximately [tex]\( 0.88 \)[/tex].
Conditional probability is the likelihood of an event occurring given that another event has already occurred. In this case, we are interested in the probability that a child spends less than 1 hour per day on electronics (Event A), given that the child spends at least 1 hour per day outside (Event B).
The formula for conditional probability [tex]\( P(A|B) \)[/tex] is given by:
[tex]\[ P(A|B) = \frac{P(A \cap B)}{P(B)} \][/tex]
Where:
- [tex]\( P(A \cap B) \)[/tex] is the probability that both Event A and Event B occur.
- [tex]\( P(B) \)[/tex] is the probability that Event B occurs.
From the given table, the total number of children surveyed is 64. Among them, the number of children who spend at least 1 hour per day outside is 16. Therefore, [tex]\( P(B) \)[/tex] is the probability that a child spends at least 1 hour per day outside.
To find [tex]\( P(A \cap B) \)[/tex]:
- We observe that there are 14 children who spend less than 1 hour per day on electronics and also spend at least 1 hour per day outside.
So:
[tex]\[ P(A \cap B) = \frac{\text{Number of children who spend less than 1 hour per day on electronics and spend at least 1 hour per day outside}}{\text{Total number of children}} = \frac{14}{64} \][/tex]
[tex]\[ P(B) = \frac{\text{Number of children who spend at least 1 hour per day outside}}{\text{Total number of children}} = \frac{16}{64} \][/tex]
Note that in conditional probability, instead of the overall total, we consider the total number of children who meet the condition (Event B). Therefore, the actual calculation involves the numbers directly:
[tex]\[ P(A|B) = \frac{14}{16} \][/tex]
By performing the division, we get:
[tex]\[ P(A|B) = \frac{14}{16} = 0.875 \][/tex]
Finally, rounding this probability to the nearest hundredth:
[tex]\[ P(A|B) \approx 0.88 \][/tex]
So, the probability that a child spends less than 1 hour per day on electronics given that the child spends at least 1 hour per day outside is approximately [tex]\( 0.88 \)[/tex].
