Tina randomly selects two distinct
numbers from the set {1, 2, 3, 4, 5}, and
Sergio randomly selects a number from the set {1, 2, . . . , 10}. The probability that Sergio’s number is larger than the sum of the two numbers chosen by Tina is ?
the number of pairs that Tina could choose is C(5,2) = 10
For each of these, Sergio can choose 10 different numbers
So we have 100 possible ways for Tina to choose 2, and Sergio to choose one number.
We can actually list them
1 2 sum of digits = 3 leaving Sergio to choose 7 larger: prob = 7/100
1 3 sum of digits = 4 leaving Sergio to choose 6 larger: prob = 6/100
1 4 sum of digits = 5 leaving Sergio to choose 5 larger: prob = 5/100
1 5 sum of digits = 6 leaving Sergio to choose 4 larger: prob = 4/100
2 3 sum of digits = 5 leaving Sergio to choose 5 larger: prob = 5/100
2 4 sum of digits = 6 leaving Sergio to choose 4 larger: prob = 4/100
2 5 sum of digits = 7 leaving Sergio to choose 3 larger: prob = 3/100
3 4 sum of digits = 7 leaving Sergio to choose 3 larger: prob = 3/100
3 5 sum of digits = 8 leaving Sergio to choose 2 larger: prob = 2/100
4 5 sum of digits = 9 leaving Sergio to choose 1 larger: prob = 1/100
sum of those probs = 40/100 = 2/5
check my logic on this one.
To find the probability that Sergio's number is larger than the sum of the two numbers chosen by Tina, we need to consider all possible outcomes.
First, let's find the total number of outcomes. Tina can choose any two distinct numbers from the set {1, 2, 3, 4, 5}. Since order doesn't matter when choosing two numbers, we can use combinations. The total number of outcomes is given by:
Total outcomes = C(5, 2) = 10
Now, let's consider the favorable outcomes where Sergio's number is larger than the sum of Tina's two numbers. There are two cases:
1. Tina chooses the numbers 1 and 2: In this case, Sergio's number must be greater than 1 + 2 = 3. Since Sergio can choose any number from 3 to 10, there are 10 - 3 + 1 = 8 favorable outcomes.
2. Tina chooses the numbers 1 and 3: In this case, Sergio's number must be greater than 1 + 3 = 4. Again, Sergio can choose any number from 4 to 10, so there are 10 - 4 + 1 = 7 favorable outcomes.
Adding up the favorable outcomes, we get:
Favorable outcomes = 8 + 7 = 15
Finally, we can calculate the probability:
Probability = Favorable outcomes / Total outcomes
= 15 / 10
= 3/2
= 0.15
Therefore, the probability that Sergio's number is larger than the sum of the two numbers chosen by Tina is 0.15.
To find the probability that Sergio's number is larger than the sum of the two numbers chosen by Tina, we need to consider all the possible outcomes and count the favorable outcomes.
Let's start by listing all the possible outcomes for the two numbers Tina can choose. Since Tina selects two distinct numbers from the set {1, 2, 3, 4, 5}, there are a total of 5 * 4 = 20 possible outcomes.
Next, we need to consider all the possible outcomes for Sergio's number. Sergio selects a number from the set {1, 2, . . . , 10}, so there are 10 possible outcomes for him.
Now let's calculate the number of favorable outcomes, i.e., the number of outcomes where Sergio's number is larger than the sum of Tina's two numbers.
To start, consider Tina's two possible numbers: (1, 2), (1, 3), (1, 4), (1, 5), (2, 3), (2, 4), (2, 5), (3, 4), (3, 5), (4, 5).
For each pair of Tina's numbers, let's calculate the number of favorable outcomes for Sergio's number to be greater.
If Tina chooses (1, 2), Sergio's number must be greater than 1 + 2 = 3, which are numbers 4, 5, 6, 7, 8, 9, 10. So there are 7 favorable outcomes for this pair.
Similarly, for (1, 3), Sergio's number must be greater than 1 + 3 = 4, so there are 6 favorable outcomes.
For (1, 4), Sergio's number must be greater than 1 + 4 = 5, so there are 5 favorable outcomes.
For (1, 5), Sergio's number must be greater than 1 + 5 = 6, so there are 4 favorable outcomes.
For (2, 3), Sergio's number must be greater than 2 + 3 = 5, so there are 5 favorable outcomes.
Similarly, for (2, 4), (2, 5), (3, 4), and (3, 5), there are 4, 3, 3, and 2 favorable outcomes respectively.
Summing up the favorable outcomes from each pair, we have 7 + 6 + 5 + 4 + 5 + 4 + 3 + 3 + 2 = 39 favorable outcomes.
Therefore, the probability that Sergio's number is larger than the sum of Tina's two numbers is 39/200.
So, the probability is 39/200 or approximately 0.195.