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. what is the probability that the first number selected was 3 or higher, given that the sum of the two numbers was 7?
options:
a. $\frac{3}{25}$
b. $\frac{4}{25}$
c. $\frac{1}{5}$
d. $\frac{3}{4}$
(table displays a 5×5 grid with cells (1,1), (1,2), ..., (5,5))

Explanation:

Step1: Identify outcomes with sum 7

First, find all pairs \((x,y)\) where \(x + y=7\). From the table:

• \(2 + 5 = 7\) (pair (2,5))
• \(3 + 4 = 7\) (pair (3,4))
• \(4 + 3 = 7\) (pair (4,3))
• \(5 + 2 = 7\) (pair (5,2))

So there are 4 outcomes where the sum is 7.

Step2: Identify which have first number ≥3

Now, check which of these have \(x\geq3\) (first number \(x\)):

• (3,4): \(3\geq3\)
• (4,3): \(4\geq3\)
• (5,2): \(5\geq3\)
• (2,5): \(2<3\) (excluded)

So there are 3 outcomes where first number is ≥3 and sum is 7.

Step3: Calculate conditional probability

Conditional probability \(P(A|B)=\frac{P(A\cap B)}{P(B)}\). Here, \(A\) is "first number ≥3", \(B\) is "sum = 7". \(n(A\cap B) = 3\), \(n(B)=4\). So \(P=\frac{3}{4}\)? Wait, no, wait: Wait, let's re - check the pairs:

Wait, the pairs with sum 7:
Looking at the table rows (first number) and columns (second number):

• Row 2 (first number 2): column 5: (2,5) sum 7
• Row 3 (first number 3): column 4: (3,4) sum 7
• Row 4 (first number 4): column 3: (4,3) sum 7
• Row 5 (first number 5): column 2: (5,2) sum 7

So total outcomes with sum 7: 4.

Outcomes with first number ≥3 and sum 7: (3,4), (4,3), (5,2) → 3? Wait, no: (3,4): first number 3 (≥3), (4,3): first number 4 (≥3), (5,2): first number 5 (≥3), (2,5): first number 2 (<3). So 3? Wait, but wait, let's count again. Wait, (3,4), (4,3), (5,2) → that's 3? Wait, no, (3,4), (4,3), (5,2) and (2,5). Wait, no, the number of outcomes with sum 7 is 4. Now, how many of these have first number ≥3?

First number in (3,4) is 3 (≥3), (4,3) is 4 (≥3), (5,2) is 5 (≥3), (2,5) is 2 (<3). So 3? Wait, but wait, maybe I made a mistake. Wait, let's list all pairs with sum 7:

1. (2,5)
2. (3,4)
3. (4,3)
4. (5,2)

Now, first number ≥3: (3,4), (4,3), (5,2) → 3 outcomes. So the conditional probability is the number of favorable outcomes (first number ≥3 and sum 7) divided by the number of outcomes with sum 7. So \(\frac{3}{4}\)? Wait, but let's check the options. Option D is \(\frac{3}{4}\). Wait, but let's re - check:

Wait, the formula for conditional probability is \(P(\text{first} \geq3|\text{sum}=7)=\frac{\text{Number of pairs with sum }7\text{ and first } \geq3}{\text{Number of pairs with sum }7}\)

Number of pairs with sum 7: Let's count from the table:

Row 1 (first number 1): 1 + y =7 → y=6, not in 1 - 5. So no.

Row 2 (first number 2): 2 + 5 =7 → (2,5)

Row 3 (first number 3): 3 + 4 =7 → (3,4)

Row 4 (first number 4): 4 + 3 =7 → (4,3)

Row 5 (first number 5): 5 + 2 =7 → (5,2)

So total 4 pairs.

Number of pairs with sum 7 and first number ≥3: (3,4), (4,3), (5,2) → 3.

So probability is \(\frac{3}{4}\).

Answer:

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