one of the letters from the word AMBIDEXTROUS is chosen at random. what is the probability that the chosen letter is E?
Answers
Answered by
25
total no of letters = 12
no of E in the word =1
so
probability =1/12
hope this helps u :-)
no of E in the word =1
so
probability =1/12
hope this helps u :-)
Anonymous:
bc 14 h
na ki 12
andha h kya be
its correct thanks ya
welcome
Answered by
7
the probability is 1/12
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