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The letters of the word ‘PROBABILITY’ are arranged in all possible ways. The chance that two B’s and also two I’s occur together is
- A\(\frac{1}{{55}}\)
- B\(\frac{2}{{55}}\)
- C\(\frac{4}{{165}}\)
- DCannot be Determined
Answer not set
Discussion 1
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CCommunity
? Probability = [No of Favorable Outcomes /No. of Total Outcomes]
⇒ No. of Total Outcomes = Total Ways of Arranging word "PROBABILITY"
Number of ways of arranging ‘n’ things in which ‘p’ are of one type ‘q’ are of one type and rest are different = $(\frac{{n!}}{{p! \times q!}})$
= Arranging a 11 letter word where two letters occur Twice
$(= \frac{{11!}}{{2! \times 2!}} = \frac{{11!}}{4})$
Assume two B as a one unit and two ‘I’ as another unit, So Number of favourable cases in which two ‘B’ together and 2 ‘I’ are together = 9!
⇒ Probability $(= \frac{{9!}}{{\frac{{11!}}{4}}})$
$(= \frac{4}{{10 \times 11}})$
= 2/55
⇒ No. of Total Outcomes = Total Ways of Arranging word "PROBABILITY"
Number of ways of arranging ‘n’ things in which ‘p’ are of one type ‘q’ are of one type and rest are different = $(\frac{{n!}}{{p! \times q!}})$
= Arranging a 11 letter word where two letters occur Twice
$(= \frac{{11!}}{{2! \times 2!}} = \frac{{11!}}{4})$
Assume two B as a one unit and two ‘I’ as another unit, So Number of favourable cases in which two ‘B’ together and 2 ‘I’ are together = 9!
⇒ Probability $(= \frac{{9!}}{{\frac{{11!}}{4}}})$
$(= \frac{4}{{10 \times 11}})$
= 2/55