Question

A six-sided number cube is “weighted” so that the numbers do not all have the same probability of landing up. The probability model is shown. Outcome 1 2 3 4 5 6 Probability 1 /12 2/ 12 /2 12 /2 /12 2 12 3 12 If the number cube is tossed once, what is P(number greater than 3)?
A. 1 /2

B. 7 /12

C. 3/4

D. 5 /12

Answer 1

Answer:

P(number greater than 3) = $\frac{7}{12}$

Step-by-step explanation:

The  Probability of  getting a number greater than 3  means the possibility of getting 4  or 5 or 6

That is

P(number greater than 3) =  P(getting 4)  +  P(getting 5) + P(getting 6)

On substituting the values from the table ,

we get

P(number greater than 3)  = $\frac{2}{12} + \frac{2}{12} + \frac{3}{12}$

P(number greater than 3) = $\frac{7}{12}$

Thus if the number cube is tossed once, P(number greater than 3) is  $\frac{7}{12}$