Jim has a special keyboard with the alphabet and ten numbers. Jim is blindfolded and does not know the position of the keys. Determine the probablility if two keys are pressed and a number is pressed first and a letter second.
There are 36 keys and the first strike of the keys has a 10/36 chance of being a number. On the second strike, he could hit any of the 36 keys, but only 26 are letters. So the answer is
10/36 x 25/36 = 0.1929
Make that 10/36 x 26/36 = 0.2006
Just a thought... if Jim holds down the first key while he presses the second, then he's only got 35 to choose from for his second press, which would slightly increase his chance of hitting a letter. But I don't think that's what the question meant.
To determine the probability of pressing a number first and then a letter on Jim's special keyboard, we need to consider the total number of keys available and the number of possible outcomes that satisfy the given condition.
Let's break down the problem into two steps:
Step 1: Probability of pressing a number first
Since Jim's keyboard has 10 numbers and a total of 26 (alphabet) + 10 (numbers) = 36 keys, the probability of pressing a number first is 10/36.
Step 2: Probability of pressing a letter second
After pressing a number, there are now 35 keys remaining on the keyboard (since one number has been used), and 26 of them are letters. Therefore, the probability of pressing a letter second is 26/35.
To find the overall probability of pressing a number first and a letter second, we multiply the probabilities from the two steps:
Probability = (Number of choices for step 1 / Total number of keys) * (Number of choices for step 2 / Total remaining keys)
Probability = (10/36) * (26/35)
Simplifying this expression, we get:
Probability = 260 / (36 * 35)
Therefore, the probability that a number is pressed first and a letter is pressed second on Jim's special keyboard is 260 / 1260, which can be further simplified to 13 / 63.