1.

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

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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