A password consists of 4 alphabet letters and 4 numbers. Calculate the following two probabilities:
- $p_1$: the probability that the letters are all equal and that the numerical part contains one eight.
- $p_2$: the probability that the password has 3 numbers followed by 4 letters.
Although it sounds like an easy question, but how would I apply the definition of permutations and combinations here? Here is how I thought of solving it.
$p_1= (1/21)^4*(1/10)*(9/10)^2$
Do I need to calculate all the possible combinations here?
$p_2= 1/\binom{7}{7}=1/7!/(7-7)!= 1/7!$
since we are considering only one case among a permutation of 7 elements over 7 places.
self-study
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