Question

a computer is programmed to select two random numbers from 1 to 5. all possible outcomes are shown in the table. each outcome has equal probability.

1,1	1,2	1,3	1,4	1,5
2,1	2,2	2,3	2,4	2,5
3,1	3,2	3,3	3,4	3,5
4,1	4,2	4,3	4,4	4,5
5,1	5,2	5,3	5,4	5,5

what is the probability that the first number selected was 3 or higher, given that the sum of the two numbers was 7?

a. $\frac{3}{25}$

b. $\frac{4}{25}$

c. $\frac{1}{5}$

d. $\frac{3}{4}$

Explanation:

Step1: Find outcomes with sum 7

First, identify all pairs \((a,b)\) where \(a + b=7\) from the table. The pairs are: \((2,5)\) (since \(2 + 5 = 7\)), \((3,4)\) (since \(3+4 = 7\)), \((4,3)\) (since \(4 + 3=7\)), \((5,2)\) (since \(5+2 = 7\)). So there are 4 such outcomes.

Step2: Find outcomes with sum 7 and first number ≥3

Now, from these 4 outcomes, find those where the first number \(a\geq3\). The pairs are \((3,4)\), \((4,3)\), \((5,2)\). So there are 3 such outcomes.

Step3: Calculate conditional probability

The formula for conditional probability \(P(A|B)=\frac{P(A\cap B)}{P(B)}\). Here, event \(A\) is "first number ≥3" and event \(B\) is "sum is 7". Since \(A\cap B\) is the outcomes in both \(A\) and \(B\), and \(P(B)\) is the number of outcomes in \(B\) divided by total outcomes, but for conditional probability with equally likely outcomes, we can use the number of favorable outcomes: \(P(A|B)=\frac{\text{Number of outcomes in }A\cap B}{\text{Number of outcomes in }B}\). So that's \(\frac{3}{4}\).

Answer:

D. \(\frac{3}{4}\)