Tom says that he needs 6 rolls to obtain each possible outcome on a 6-sided die. on the fourth roll, he rolls his second "3”. tom says that the die is loaded and that each outcome is not equally likely. is tom correct here? if you think tom is incorrect, how many rolls should tom make until he sees each number occurring about 1/6 of the time?

Tom is wrong.

Step-by-step explanation:

Tom is wrong.

There are 6 possible outcomes when rolling a dice {1, 2, 3,4, 5, 6}

The probability of obtaining one of these numbers is always 1/6 in each trial

That is, the probability of obtaining a number does not depend on the result obtained in the previous trial. Therefore it is possible to obtain the same number more than once.

We know that the larger the number of trials, the more likely it is to obtain each die number 1/6 of the time.

This happens because we know that the probability of getting each face on the dice is 1/6. Then the experimental probability must converge to the same value as the theoretical probability when the number of trials tends to infinity.

This means that: the more trials you do, the more you will get to get each face 1/6 of the time.

x=-7

step-by-step explanation:

x+3=-10

-10+3=-7

x=-7