Answer:
(a) 0.25
(b) 0.944
(c) 0.444
(d) 0.167
Step-by-step explanation:
There are six possible outcomes for each die, which means that the number of possible outcomes is:
[tex]n=6*6 = 36[/tex]
(a) In order for each die to show four or more spots they will both have to land on a four, five or six. The probability of this happening is:
[tex]P(A) = \frac{3*3}{36}=0.25[/tex]
(b) There are only two possible outcomes for which the sum is three (1 and 2, or 2 and 1). The probability of the sum NOT being three is:
[tex]P(B) = 1-\frac{2}{36}=0.944[/tex]
(c) If neither a one or a six must appear, then there are 4 possible outcomes for each die, the probability is:
[tex]P(C) = \frac{4*4}{36}=0.444[/tex]
(d) For each one of the six possible numbers on the first die, there is only one on the second die for which the sum of the spots is 7, totaling six possible ways to sum 7:
[tex]P(D) = \frac{6}{36}=0.167[/tex]
The required probability output from a throw of two six sided dice are as follows :
The sample space for two throw of a six-sided die :
Recall :
A.) Obtaining 4 or more spots :
Required spot = (5, 6, 7) on each die = 3 × 3 = 9 outcomes
P(4 or more spot) = 9/36 = 0.25
B.) Sum of spot is not 3 :
Sum of spot = 3 ; (1, 2) and (2, 1) = 2 possible outcomes
P(sum not 3) = 1 - (2/36) = 1 - 1/8 = 17/18 = 0.944
C.) neither a one nor 6 appears :
Required = (2, 3, 4, 5) = 4 × 4 = 16
P(neither 6 nor 1) = 16/36 = 4/9 = 0.44
D.) Sum of spot equals 7
Required = (1, 6),(6,1),(5,2),(2,5),(3,4),(4,3) = 6 outcomes
P(sum equals 7) = 6/36 = 1/6 = 0.167
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