Solve the problem.
-A box contains 4 slips of paper, on each of which is written the number 1, 2, 3, or 4, respectively. A slip is drawn, and the number on it is noted. That slip is put aside, and another slip is drawn and the number on it
Noted. What is the probability that
1) the sum of the two numbers is 5?
2) the first number drawn is a 2 and the second number is not 5?
3) the sum of the two numbers is not 10?
A) a) $\frac { 1 } { 4 }$ b) $\frac { 1 } { 3 }$ c) 0

B) a) $\frac { 1 } { 3 }$ b) $\frac { 1 } { 4 }$ c) 0

C) a) $\frac { 1 } { 4 }$ b) $\frac { 1 } { 3 }$ c) 1
D)a) $\frac { 1 } { 3 }$ b) $\frac { 1 } { 4 }$ c) 1

Question

Solve the problem.
-A box contains 4 slips of paper, on each of which is written the number 1, 2, 3, or 4, respectively. A slip is drawn, and the number on it is noted. That slip is put aside, and another slip is drawn and the number on it
Noted. What is the probability that
1) the sum of the two numbers is 5?
2) the first number drawn is a 2 and the second number is not 5?
3) the sum of the two numbers is not 10? 
A) a) $\frac { 1 } { 4 }$ b) $\frac { 1 } { 3 }$ c) 0

B) a) $\frac { 1 } { 3 }$ b) $\frac { 1 } { 4 }$ c) 0

C) a) $\frac { 1 } { 4 }$ b) $\frac { 1 } { 3 }$ c) 1
D)a) $\frac { 1 } { 3 }$  b) $\frac { 1 } { 4 }$ c) 1

Quizplus · Accepted Answer

1) The pairs that sum to 5 are (1,4), (2,3), (3,2), and (4,1), so 4 out of 12 possible outcomes, giving $\frac{1}{3}$. 2) Drawing a 2 first has a $\frac{1}{4}$ chance, and since the second number cannot be 5 (not in the box), this condition doesn't change the probability. 3) Since the numbers only go up to 4, it's impossible for their sum to be 10, so the probability of not summing to 10 is 1.

Solve the Problem 14\frac { 1 } { 4 }41​ B) 13\frac { 1 } { 3 }31​

Solve the Problem $\frac { 1 } { 4 }$ B) $\frac { 1 } { 3 }$