Question

The two-way table shows the number of houses on the market in the Castillos’ priceA 5-column table has 4 rows. The first column has entries A, B, C, Total. The second column is labeled X with entries 15, 5, 30, 50. The third column is labeled Y with entries 5, 8, 15, 28. The fourth column is labeled Z with entries 10, 7, 5, 22. The fifth column is labeled Total with entries 30, 20, 50, 100. Which two events are independent? A and X A and Y B and X B and Y range. A 6-column table has 4 rows. The first column has entries 1 bathroom, 2 bathrooms, 3 bathrooms, total. The second column is labeled 1 bedroom with entries 67, 0, 0, 67. The third column is labeled 2 bedrooms with entries 21, 6, 18, 45. The fourth column is labeled 3 bedrooms with entries 0, 24, 16, 40. The fifth column is labeled 4 bedrooms with entries 0, 0, 56, 56. The sixth column is labeled Total with entries 88, 30, 90, 208. What is the probability that a randomly selected house with 2 bathrooms has 3 bedrooms? 0.2 0.4 0.6 0.8

Answer 1

To determine which two events are independent in the given two-way table, we need to check whether the probability of one event is affected by the occurrence of the other event. If the probability of one event is not affected by the occurrence of the other event, then the two events are independent.

Looking at the table, we can see that the total number of houses on the market is 100. The number of houses in category A is 30, and the number of houses in category X is 50. So, the probability of a house being in category A is 30/100 or 0.3, and the probability of a house being in category X is 50/100 or 0.5.

Similarly, the number of houses in category A and Y is 5, so the probability of a house being in both categories A and Y is 5/100 or 0.05. The number of houses in category A and Z is 10, so the probability of a house being in both categories A and Z is 10/100 or 0.1.

Now, to check whether events A and X are independent, we need to compare the probability of a house being in category A with the probability of a house being in category A given that it is also in category X. The probability of a house being in category A given that it is also in category X is 0.15 (i.e., 15/50), which is the numberof houses in both categories A and X divided by the total number of houses in category X. Since the probability of a house being in category A (0.3) is not affected by the occurrence of it being in category X, events A and X are independent.

Similarly, we can check whether events A and Y, B and X, and B and Y are independent. The probabilities of a house being in categories A and Y, B and X, and B and Y are 0.05, 0.1, and 0.14, respectively. The probability of a house being in category A given that it is also in category Y is 5/28, which is not equal to the probability of a house being in category A (0.3). Therefore, events A and Y are not independent. Similarly, events B and X and B and Y are not independent either.

Hence, the two events that are independent are A and X.

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To calculate the probability that a randomly selected house with 2 bathrooms has 3 bedrooms from the given table, we need to look at the row corresponding to 2 bathrooms, and then find the probability of a house in that row having 3 bedrooms. The number of houses with 2 bathrooms and 3 bedrooms is 18, out of a total of 30 houses with 2 bathrooms. Therefore, the probability of a randomly selected house with 2 bathrooms having 3 bedrooms is 18/30, which simplifies to 0.6.

So, the answer is:

0.6.