Problem 21 - Algebra 2

Question

The tiles \(E, T, F, U, N, X,\) and \(P\) of a word game are placed face down in the lid of the game. If two tiles are chosen at random, find each probability. \(P(X, \text { then } P),\) if no replacement occurs

Step-by-Step Solution

Answer

The probability is \(\frac{1}{42}\).

Step 1: Determine Total Possible Outcomes

There are 7 tiles in total: \(E, T, F, U, N, X,\) and \(P\). When selecting 2 tiles without replacement, we calculate the number of unique pairs (combinations) as choosing 2 out of 7, which is \(\binom{7}{2} = 21\).

Step 2: Calculate the Probability of Drawing X First

To find the probability of drawing \(X\) first, note that there is only 1 \(X\) tile among 7 tiles. Thus, the probability is \(\frac{1}{7}\).

Step 3: Calculate the Probability of Drawing P Second

If \(X\) is drawn first, there are now 6 tiles left: \(E, T, F, U, N, P\). The probability of drawing \(P\) from these 6 tiles is \(\frac{1}{6}\).

Step 4: Compute the Combined Probability

The probability of both events occurring in sequence (drawing \(X\) then \(P\) without replacement) is the product of their individual probabilities: \(\frac{1}{7} \times \frac{1}{6} = \frac{1}{42}\).

Key Concepts

Combinations

In probability, combinations are used to determine how many ways we can choose items from a group, where the order does not matter. Imagine you have a box of different toys, and you can only pick two at a time. Each unique combination of toys you pick is counted once, regardless of the order you pick them in.

For example, in our exercise, there are 7 tiles labeled as: E, T, F, U, N, X, and P. We're interested in picking any 2 of them out, whether it's EF or FE, both count as the same combination.

To calculate this, we use the combination formula, which is: \[\binom{n}{r} = \frac{n!}{r!(n-r)!}\]where \( n \) is the total number of items, and \( r \) is the number of items to choose. In this case, it simplifies to choosing 2 tiles out of 7, leading to: \[\binom{7}{2} = 21\]This calculation tells us there are 21 possible ways to choose 2 tiles from the group of 7.

Event sequence

When dealing with probabilities involving sequences, order becomes important. An event sequence is a series of occurrences that happen one after the other. Each step in the sequence may influence the following ones. In our tile exercise, we want to find the probability of drawing tile X followed by tile P.

Event sequences often need you to calculate probabilities step-by-step. In our case:

• First, consider picking tile X. The chance of this happening depends on how many tile X's there are in the group. With just one X out of 7 tiles, it's a probability of \(\frac{1}{7}\).
• Then, if X is picked, there’s now one less tile in the pool, leaving 6 tiles. Among these, you want to know the chance of picking tile P, which would be \(\frac{1}{6}\) since there’s one P among 6 tiles.

Events in a sequence mean the outcome of one can change the probability of the next. Multiply the probabilities of each step to get the sequence's total probability, as we did by finding \(\frac{1}{7} \times \frac{1}{6} = \frac{1}{42}\).

Without replacement

When items are drawn without replacement, it means once an item is picked, it is not put back into the pool of choices. This impacts the calculations of probabilities because the total number of items changes with each draw. In our exercise, once you draw a tile, it stays out—so every subsequent draw has one less tile to choose from.

Focusing on our example, let’s see how this affects our calculations:

• Initially, with 7 tiles, the probability of picking the tile X is \(\frac{1}{7}\).
• Once X is picked, it’s set aside: hence, for the next draw, you don't replace X back into the tile set.
• This leaves us with only 6 tiles for the next choice, affecting the probability of picking tile P to be \(\frac{1}{6}\).

Without replacement is a crucial concept because it ensures probabilities are correctly adjusted for subsequent choices, demonstrating how interconnected events can be affected by previous ones.