Question

5 red balls and 6 black balls are in a bag. Find the probability that in two draws a red
ball and a black ball are chosen. No replacement of names is done after a draw

Answer 1

1/11 is the probability that when u pick the ball at random it will either be a red one or black one

Answer 2

but answer is 6/11

Answer 3

Okay, there are two ways the draw can happen. Either a red ball can be drawn first or a black ball can be drawn first.

Answer 4

6/11 is the probability of picking up a black ball only

Answer 5

If the first draw that happens is a red ball and the second draw is a black ball, then the probability is calculated as follows:
\[\frac{ 5 }{ 11 } \times \frac{ 6 }{ 10 }\]which simplifies to\[\frac{ 5 }{ 11 } \times \frac{ 3 }{ 5 }\]

Answer 6

@ahtesham No, it isn't.

Answer 7

\[\frac{ 5 }{ 11 } \times \frac{ 3 }{ 5 } = \frac{ 15 }{ 55 } = \frac{ 3 }{ 11 }\]

Answer 8

Now if the first draw is a black ball, and then second draw is a red ball, the probability is again calculated as follows: \[\frac{ 6 }{ 11 } \times \frac{ 5 }{ 10 }\]which simplifies to\[\frac{ 6 }{ 11 } \times \frac{ 1 }{ 2 }\]

Answer 9

\[\frac{ 6 }{ 11 } \times \frac{ 1 }{ 2 } = \frac{ 6 }{ 22 } = \frac{ 3 }{ 11 }\]

Answer 10

Now we take the two possible probabilities of drawing a combination of a red ball and a black ball and we add them together to get the final probability that such a combination will be drawn.

Answer 11

Therefore, the final probability is\[\frac{ 3 }{ 11 } + \frac{ 3 }{ 11 } = \frac{ 6 }{ 11 }\]

Answer 12

thanks

Answer 13

Did you understand all that? xD