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Standard XII

Mathematics

Multiplication Rule

Seven white b...

Question

Seven white balls and three black balls are randomly placed in a row. The probability that no two black balls are placed adjacently, equals?

\frac{1}{2}

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\frac{7}{15}

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\frac{2}{15}

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\frac{1}{3}

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Solution

The correct option is B $\frac{7}{15}$

Total number of ways in which all the balls can be placed is
10!/(7!*3!)=120
If no two black balls are placed adjacently, the possible number of ways is
${8_{C}}_{3}$ =56
Hence the required chance is = 56/120=7/15

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Seven white balls and three black balls are randomly placed in a row. The probability that no two black balls are placed adjacently, equals?

The correct option is B 715Total number of ways in which all the balls can be placed is10!/(7!*3!)=120=120If no two black balls are placed adjacently, the possible number of ways is8C3=568C3=56Hence the required chance is = 56/120=7/1556120

The correct option is B $\frac{7}{15}$

Total number of ways in which all the balls can be placed is
10!/(7!*3!)=120
If no two black balls are placed adjacently, the possible number of ways is
${8_{C}}_{3}$ =56
Hence the required chance is = 56/120=7/15