There are 10 letters in which A, B, and L each
appear twice.
The number of distinguishable permutations is
10 / 2! x 2! x 2! = 453,600
The total number of possible outcomes is 453,600 and
there is only one favorable outcome which is
BASKETBALL. Therefore, the probability is
ANSWER: 1/453,600
Answer:
[tex]\dfrac{1}{453600}[/tex]
Step-by-step explanation:
To calculate the probability of forming the word "BASKETBALL" from the permutation of the letters B-A-S-K-E-T-B-A-L-L, we need to find out how many permutations of these letters are possible and how many of them result in the word "BASKETBALL".
First, determine the total number of permutations of the letters in "BASKETBALL". Since there are 10 letters in total, and some letters repeat (2 B's, 2 A's, and 2 L's), we need to account for that using the permutations with repetition formula:
[tex]\textsf{Total permutations} = \dfrac{10!}{2! \times 2! \times 2!}\\\\\\\textsf{Total permutations} = \dfrac{10\times 9\times 8\times 7\times 6\times 5\times 4\times 3\times 2\times 1}{2 \times 1 \times 2 \times 1 \times2 \times 1}\\\\\\\textsf{Total permutations} = \dfrac{10\times 9\times 8\times 7\times 6\times 5\times 4\times 3}{2 \times 1 \times 2 \times 1}\\\\\\\textsf{Total permutations} = \dfrac{1814400}{4}\\\\\\\textsf{Total permutations} =453600[/tex]
As there is only one arrangement that forms the word "BASKETBALL", the probability of forming the word "BASKETBALL" is:
[tex]\dfrac{1}{453600}[/tex]
So, the probability that the randomly selected permutation forms the word "BASKETBALL" is:
[tex]\Large\boxed{\boxed{\dfrac{1}{453600}}}[/tex]