Question

There are two spinners. The first spinner has three equal sectors labeled 1, 2 and 3. The second spinner has four equal sectors labeled 3, 4, 5 and 6. The spinners are spun once. What is the number of possible outcomes that do not show a 1 on the first spinner and show the number 4 on the second spinner?
2
6
9
12

Answer 1

The number of possible outcomes that do not show a 1 on the first spinner and show the number 4 on the second spinner is 2.

Spinner 1 Spinner 2
1                   1
1                   2
1                   3
1                   4
2                  1
2                  2
2                  3
2                  4  = NOT 1, AND 4
3                  1
3                  2
3                  3
3                  4 = NOT 1, AND 4

Only 2 outcomes have passed the given condition.

Answer 2

Answer:  First option is correct.

Step-by-step explanation:

Since we have given that

the first spinner has three equal sectors labelled 1, 2 and 3

The second spinner has four equal sectors labelled 3, 4, 5 and 6.

Sample space will be

${(1,3),(1,4),(1,5),(1,6),(2,3),(2,4),(2,5),(2,6),(3,3),(3,4),(3,5),(3,6)}$

So, Number of possible outcomes that do not show a 1 on the first spinner and show the number 4 on the second spinner :

As we can see that there are 2 such outcomes where we can get the above condition i.e. (2,4) and (3,4).

Hence, First option is correct.