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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 is1). \(\frac{1}{{55}}\)2). \(\frac{2}{{55}}\)3). \(\frac{4}{{165}}\)4). Cannot be Determined |
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Answer» ? 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 |
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