Question

Two spinners with three equal sections are spun.

Each spinner is spun at the same time and their results are added together.

One is labeled with the numbers 1, 2, and 3. The other is labeled with the numbers 4, 5, and 6.

About how many times would you expect to spin a sum of 7 out of 100 spins? ​

Answer 1

out of a 100 spins, we expect 33.33 to give a sum of 7

So, rounding to the nearest whole number, we expect to get a sum of 7 33 times out of a hundred

Explanation:

We note that 1+6 = 7, 2+5 = 7, 3+4 = 7,

Now, since the sections of the spinners are equal,

The probability that they stop at any number is 1/3 (since there are 3 sections)

, now, for, 1+6, the 1st spinner stops at 1, and the 2nd spinner stops at 6,

The probability of this happening is,

(1/3)(1/3) = 1/9

Similarly for 2+5 we get, (1/3)(1/3) = 1/9

And for 3+4, the 1st spinner stops at 3, and the 2nd spinner stops at 4,

The probability is,

(1/3)(1/3) = 1/9

So, the total probability that the sum is 7 is,(for a single try) the sum of these probabilities,

P = either the sum is 1+6 or 2+5 or 3+4,

P = 1/9 + 1/9 + 1/9 = 3/9

P = 1/3

For 1 try, the chance is 1/3, for 100 tries, we multiply this by 100,

(1/3)(100) = 33.33

So, out of a 100 spins, we expect 33.33 to give a sum of 7 or, 33-34 will give a sum of 7