A bag contains 6 black- 5 white balls- One ball is drawn randomly- The probability of it being a black ball is-dfrac-6-11-dfrac-5-11-dfrac-1-11-dfrac-27-128-

Question

Toppr Admin · Accepted Answer

The correct option is A 611Total number of balls in the bag- n-S-6-5-11Hence the probability of it being a black ball is- n-E-n-S-611

A bag contains $6$ black, $5$ white balls. One ball is drawn randomly. The probability of it being a black ball is......
$\frac{6}{11}$
$\frac{5}{11}$
$\frac{1}{11}$
$\frac{27}{128}$

The correct option is A $\frac{6}{11}$
Total number of balls in the bag, $n (S) = 6 + 5 = 11$
Hence the probability of it being a black ball is $=$ $\frac{n (E)}{n (S)} = \frac{6}{11}$

A bag contains 6 black, 5 white balls. One ball is drawn randomly. The probability of it being a black ball is......61151111127128

The correct option is A 611Total number of balls in the bag, n(S)=6+5=11Hence the probability of it being a black ball is= n(E)n(S)=611

A bag contains $6$ black, $5$ white balls. One ball is drawn randomly. The probability of it being a black ball is......
$\frac{6}{11}$
$\frac{5}{11}$
$\frac{1}{11}$
$\frac{27}{128}$

The correct option is A $\frac{6}{11}$
Total number of balls in the bag, $n (S) = 6 + 5 = 11$
Hence the probability of it being a black ball is $=$ $\frac{n (E)}{n (S)} = \frac{6}{11}$