Probability Of Forming Words: RAKETA Card Puzzle

Hey guys! Let's dive into a fun probability puzzle involving the letters of the word RAKETA. This is a classic problem that tests our understanding of permutations, combinations, and the fundamental principles of probability. We'll break it down step-by-step, so you can see exactly how to tackle these kinds of problems. So, grab your thinking caps, and let's get started!

Part A: Probability of Drawing REKA

In this first part, we're figuring out the odds of drawing the letters R, E, K, and A, in that specific order, from our set of six cards. This involves a bit of combinatorial thinking and a solid grasp of probability basics. Let's break it down, guys.

When calculating probability, we always start by considering the total possible outcomes. In this case, we are drawing four letters out of six. The order matters here because we want to form the word REKA. This means we're dealing with permutations. The total number of ways to pick and arrange 4 letters from 6 is given by the permutation formula:

P(n, r) = n! / (n - r)!

Where n is the total number of items (6 letters in RAKETA), r is the number of items we're choosing (4 letters), and ! denotes the factorial (e.g., 5! = 5 * 4 * 3 * 2 * 1). So, plugging in our numbers:

P(6, 4) = 6! / (6 - 4)! = 6! / 2! = (6 * 5 * 4 * 3 * 2 * 1) / (2 * 1) = 360

Therefore, there are 360 different ways to draw four letters from the six letters in RAKETA if we care about the order.

Now, let's think about the favorable outcomes. We want to draw the letters R, E, K, and A in that exact order. There's only one way to do this. We need to pick R first, then E, then K, and finally A.

Probability is defined as the number of favorable outcomes divided by the total number of possible outcomes. So, the probability of drawing REKA is:

Probability (REKA) = (Number of favorable outcomes) / (Total number of possible outcomes) = 1 / 360

So, the probability of drawing the letters R, E, K, and A in the correct order to form the word REKA is a rather slim 1 in 360. It highlights how specific arrangements become less likely as the number of possibilities grows. Imagine shuffling a deck of cards and trying to get a specific four-card sequence – it's a similar kind of rarity!

Part B: Probability of Forming KARETA

Okay, guys, let's crank up the complexity a little bit! Now we're tasked with figuring out the likelihood of forming the word KARETA when drawing all six letters. This twists things slightly because we need to account for repeated letters. The presence of two 'A's changes the game, making us carefully consider how we count arrangements.

Firstly, let’s calculate the total possible outcomes. We have six letters, but since the letter 'A' appears twice, we can't just calculate 6! (which would be 6 * 5 * 4 * 3 * 2 * 1 = 720) as the total number of permutations. This is because swapping the two 'A's doesn't create a new distinct arrangement. To account for this repetition, we use the formula for permutations with repetitions:

Total permutations = n! / (n1! * n2! * ... * nk!)

Where n is the total number of items (6 letters), and n1, n2, ..., nk are the counts of each repeated item. In our case, n = 6, and we have one repeated letter 'A' which appears twice (so n1 = 2). Thus:

Total permutations = 6! / 2! = (6 * 5 * 4 * 3 * 2 * 1) / (2 * 1) = 720 / 2 = 360

So, there are 360 distinct ways to arrange the letters of the word RAKETA.

Next, we need to determine the favorable outcomes. How many ways can we arrange the letters to spell KARETA? Well, there's only one way to spell KARETA. The letters must be in that exact order.

Therefore, the probability of forming the word KARETA is:

Probability (KARETA) = (Number of favorable outcomes) / (Total number of possible outcomes) = 1 / 360

Interestingly, the probability of forming KARETA is the same as the probability of drawing REKA in the first part of the question! This might seem counterintuitive at first, given the different scenarios. However, it highlights how the constraints of the problem – the specific arrangement we're looking for versus the total possible arrangements – dictate the final probability. Whether we're drawing four letters or arranging all six, the rarity of a single, specific sequence amidst many possibilities results in the same probability in this case.

Key Takeaways

This problem really drives home some crucial concepts in probability and combinatorics:

• Permutations: The order matters! When we're arranging items and the sequence is important, we use permutations.
• Combinations: The order doesn't matter! If we're just selecting items without regard to their arrangement, we use combinations.
• Factorials: Factorials (like 5!) are the bread and butter of counting arrangements. They represent the product of all positive integers up to a given number.
• Repetitions: Repeated items (like the two 'A's in RAKETA) require special handling when calculating permutations. We need to divide by the factorial of the count of each repeated item to avoid overcounting.
• Probability as a Ratio: Probability is fundamentally the ratio of favorable outcomes to total possible outcomes. Keep this definition in your mind, and you'll have a solid foundation for tackling probability problems.

Final Thoughts

Probability problems can be a real brain-bender, but they're also super rewarding! By breaking down the problem into smaller steps – identifying the total outcomes, figuring out the favorable outcomes, and then calculating the ratio – you can conquer even the trickiest puzzles. Guys, keep practicing, and you'll be a probability pro in no time! Remember, the key is to think systematically, understand the underlying concepts, and not be afraid to experiment with different approaches. Now, go forth and solve some more probability problems!