A red ball and 6 white balls are in a box. If two balls are drawn, without replacement, what is the probability of getting a red ball on the first draw and a white ball on the second?
a. 1/6
b. 1/7
c. 6/7
d. 7/13

Question

A red ball and 6 white balls are in a box. If two balls are drawn, without replacement, what is the probability of getting a red ball on the first draw and a white ball on the second?
a. 1/6
b. 1/7
c. 6/7
d. 7/13

anonymous · Answer

You have a 1/7 chance to draw a red ball first. Since you aren't replacing the ball or putting it back into the box, all you have left are white balls in the box. So you have a 100% or 1.0 chance to get the white ball. 

So:

$$1/7 \times 1 = 1/7$$

anonymous · Answer

what about the white ball on the second draw?

anonymous · Answer

You have a 100% chance of getting the white ball. So its 1/7 multiplied by 1. The 1 is the 100%.

anonymous · Answer

1/7   the probability of red coming for the first time is 1/7 and then there are only white ones whose probability become 1

anonymous · Answer

Notation wise, it can be written as:

$$(1/7) \times (6/6)$$

anonymous · Answer

Which is equal to $$1/7$$

anonymous · Answer

oooh i see. thank you guys!