The SIG Problem-Solving Assessment - What It Is and How to Ace It

The answer is 7/3

This is a question of binominal distribution. We can think of each card draw as a trial. A successful trial is when a card with a prime number is drawn from the deck.

First, let’s compute the probability to draw a prime numbered card in a single draw. Within [1-18] there are 7 prime numbers (2,3,5,7,11,13,17). Thus, the probability to draw a prime numbered card in a single draw is

Since this is a binominal distribution, we can compute the expected number of card draws according to the following (n = number of draws):

19/200

This question combines combinatorial thinking with probability.

To find the probability that exactly two printers will have a paper jam, we need to consider all the possible scenarios in which two out of the three printers have a paper jam. Then, we will compute the sum of the probabilities of these scenarios.

Printer 1:

• P(paper jam) = 0.25
• P(no paper jam) = 0.75

Printer 2:

• P(paper jam) = 0.2
• P(no paper jam) = 0.8

Printer 3:

• P(paper jam) = 0.15
• P(no paper jam) = 0.85

Scenario 1:

Printers 1 and 2 have paper jam, and printer 3 doesn’t.

The probability for scenario 1 is:

0.25×0.2×0.85=0.04250.25×0.2×0.85=0.0425

Scenario 2:

Printers 1 and 3 have paper jam, and printer 2 doesn’t.

The probability for scenario 1 is:

0.25×0.8×0.15=0.030.25×0.8×0.15=0.03

Scenario 3:

Printers 2 and 3 have paper jam, and printer 1 doesn’t.

The probability for scenario 1 is:

0.75×0.2×0.15=0.02250.75×0.2×0.15=0.0225

Thus, the probability that exactly 2 printers will have a paper jam is the sum of probabilities of the 3 scenarios we listed:

0.0425+0.03+0.0225= 0.095= 19/200

Answer: 20

Explanation: 

First, let’s look at the facts of the question:

1. To get from A to B, the robot must move 17 units to the right and 5 units upwards = a total of 22 units.
2. The robot can only go up or right and constantly alternates its step between 2 and 3 units. Thus, it can either start with a 2-unit or 3-unit step.

Let’s see what happens when the robot starts with a 3-unit step:

3, 2, 3, 2, 3, 2, 3, 2 ->  After 8 steps, the robot has passed 20 units, leaving only 2 units to the next step. Since the next step must be a 3-unit step and the robot can’t overshoot point B, we learn that starting with a 3-unit step isn’t possible.

Let’s see what happens when the robot starts with a 2-unit step:

2, 3, 2, 3, 2, 3, 2, 3 ->  After 8 steps, the robot has passed 20 units. The next step of the robot must be a 3-unit step, which is the exact number of units remaining to reach point B.

Thus, we know the robot must start with a 3-unit step.

Now we need to determine how many different ways the robot goes from A to B according to the alternating speed pattern we found (2, 3, 2, 3, 2, 3, 2, 3, 2).

Since the robot can go upwards a total of 5 units (due to the limitations in the question), it must include one 3-unit step upwards and one 2-unit step upwards.

As we have already seen, the robot’s path consists of four 3-unit steps and five 2-unit steps.

1. For the 2-unit step upwards – the robot has five different options (since there are five 2-unit steps in its path).
2. Similarly, for the 3-unit step upwards– the robot has four different options.

Thus, the total number of possible paths that the robot can go from point A to B is 5x4=20.