Question

a bag contains 7 white , 5 black and 4 red balls if two balls are drawn random find the probabilities that i") both the balls are whiteii) both have same colours iii) both have different colours

Answer 1

do theoretical probability and count the outcomes and solve.

Answer 2

please explain

Answer 3

There are 7+5+4 = 16 balls in total

i) Both balls are white
1st ball: The probability of picking a white is 7/16
2nd ball: The probability of picking a 2nd white one is 6/15 (since one white is gone, so there are 6 white balls left, and hence 5 in total).
The probability of getting both white is thus (7/16)*(6/15)=7/40=0.175

You can alternatively think of this in terms of combinations:
The sample space consists of all 2-ball draws from the 16 balls. Since the order doesn't matter, you can do this in \(\Large {16 \choose 2}=120\) ways.

Now for the event, you are picking 2 white balls from 7 white ones, so you can do this in \(\Large {7 \choose 2}=21\) ways.

Hence the probability is:
\[\Large \frac{{7 \choose 2}}{{16 \choose 2}}=\frac{21}{120}=0.175 \]
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ii) Both are the same color. Denote W =White, B=Black, R=RedThis means you can get either:
(W, W) or (B, B) or (R, R), hence:

P(2 balls same color) = P(W, W) + P(B, B) + P(R, R)

P(W, W) is exactly what we calculated in i)

P(B, B) will have a similar argument to i) and is given by:
\[\Large \frac{{5 \choose 2}}{{16 \choose 2}} \]

P(R, R) is again similar, but taking into account that you have 4 Reds.

So the overall probability is P(W, W) + P(B, B) + P(R, R)
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iii) Both have different colors, this means you can get
(W, B) or (W, R) or (B, R)

Notice that (W, B) and (B, W) are the same combination since the order doesn't matter, so we only consider (W, B). That's why's there are only 3 combinations to look at. Again, the overall probability will be:
P(2 different colors) = P(W, B) + P(W, R) + P(B, R)

So, looking at P(W, B):
The same space is the same as in ii) (120 = (16 choose 2) ) since you are again picking 2 balls in total from 16.
1st ball: You are picking 1 ball from the 7 white ones, which is represented as \(\Large {7 \choose 1}\)
2nd ball: You are picking 1 ball from the 5 black ones, which is represented as \(\Large {5 \choose 1}\)
So, the overall probability for P(W, B) is:
\[\Large \frac{{7 \choose 1}{5 \choose 1}}{{16 \choose 2}} \]
You can proceed in the same way for P(W, R) and P(B, R)

Answer 4

i got the same answer