Solveeit Logo
Login

Question

Mathematics Question on Probability

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

A

12\frac{1}{2}

B

715\frac{7}{15}

C

1

D

815\frac{8}{15}

Answer

715\frac{7}{15}

Explanation

Solution

The number of ways of placing 3 black balls without any restriction is 10C3.^{10}C_3. Since, we have total 10 places of putting 10 balls in a row. Now, the number of ways in which no two black balls put together is equal to the number of ways of choosing 3 places m arked '-' out of eight places              W−W−W−W−W−W−W−W− \, \, \, \, \, \, \, \, \, \, \, \, \, W-W-W-W-W-W-W-W- This can be done in 8C3^8C_3 ways ∴     \therefore \, \, \, \, \, Required probability =8C310C3=8×7×610×9× 8=715\frac{^8 C_3}{^{10}C_3}=\frac{8 \times 7 \times 6}{10 \times 9 \times \, 8}=\frac{7}{15}