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
a) the sum of the two numbers is 5?
b) the first number drawn is a 2 and the second number is not 5?
c) the sum of the two numbers is not 10?
A) a) $\frac { 1 } { 3 }$ b) $\frac { 1 } { 4 }$ c) 0
B) a) $\frac { 1 } { 4 }$ b) $\frac { 1 } { 3 }$ c) 1
C) a) $\frac { 1 } { 3 }$ b) $\frac { 1 } { 4 }$ c) 1
D) a) $\frac { 1 } { 4 }$ b) $\frac { 1 } { 3 }$ c) 0

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
a) the sum of the two numbers is 5?
b) the first number drawn is a 2 and the second number is not 5?
c) the sum of the two numbers is not 10? 
A) a) $\frac { 1 } { 3 }$ b) $\frac { 1 } { 4 }$ c) 0
B) a) $\frac { 1 } { 4 }$ b) $\frac { 1 } { 3 }$ c) 1
C) a) $\frac { 1 } { 3 }$ b) $\frac { 1 } { 4 }$ c) 1
D) a) $\frac { 1 } { 4 }$ b) $\frac { 1 } { 3 }$ c) 0

Quizplus · Accepted Answer

a) The pairs that sum to 5 are (1,4), (2,3), (3,2), and (4,1). There are 4 favorable outcomes out of 12 possible outcomes (4 choices for the first draw and 3 remaining choices for the second), so the probability is $\frac{4}{12} = \frac{1}{3}$. b) If the first number drawn is a 2, there are 3 possible outcomes for the second draw (1, 3, or 4), but since the second number cannot be 5 (which is not even an option), the probability is simply the probability of drawing a 2 first, which is $\frac{1}{4}$. c) Since the numbers only go up to 4, it's impossible for any two numbers to sum to 10, making the probability of not summing to 10 equal to 1.

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

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