A bag contains 10 marbles,7 of which are black and 3 are red. three marbles are drawn at one after the other. find the probability of choosing at most two black marbles.​

Explanation:

We have these four cases or possible outcomes

Let's calculate the probability for case 4.

There are 7 black marbles out of 10 total. The probability of picking black is 7/10. If no replacement is made, then 6/9 is the probability of picking black again (subtract 1 from the numerator and denominator separately). Finally, 5/8 is the probability of getting black a third time.

The probability of getting 3 black marbles in a row is

(7/10)*(6/9)*(5/8) = (7*6*5)/(10*9*8) = 210/720 = 7/24.

That fraction 7/24 means that if you had 24 chances, then you expect about 7 of them will lead to getting three black marbles in a row (aka case 4). Therefore, 24-7 = 17 occurrences are expected where we get cases 1 through 3 occur in some fashion (pick one case only).

Notice how cases 1 through 3 encapsulate the phrasing "at most 2 black marbles" which is another way of saying "2 black marbles is the highest we can go".

So that's why the answer is 17/24.

i did it myself and i got 4.2 (b) hope its right

step-by-step explanation: