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A bag contains 5 White,7 black, 4 Red balls. Find the probability that three balls drawn at random are white.

My attempt:

The sample space for drawing 3 balls is {WWW,BBB,RRR,WBB,WRR,RBB,RWW,BRR,BWW}. So required probability is 1/9. But answer given is 1/56. Where am I wrong?

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    $\begingroup$ You draw your conclusion from expecting that the probability of drawing each ball is the same when it clearly isn't - obviously, drawing a white ball should be more likely than drawing a red ball, but less likely than drawing a black one. $\endgroup$ Commented Oct 4, 2020 at 5:32
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    $\begingroup$ Do you think the answer still would be $\frac 19$ if the bag contained $3$ white balls, $1000$ black balls, and $1000$ red balls? But if those were the numbers, your calculation would remain exactly the same. That's a clue that your analysis can't be right. $\endgroup$ Commented Oct 4, 2020 at 5:33
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    $\begingroup$ $\frac{5\times 4\times 3}{16\times 15\times 14}$ $\endgroup$
    – acat3
    Commented Oct 4, 2020 at 5:34
  • $\begingroup$ @cosmo5 its given they are taken at random all at once. Permutations wont matter. $\endgroup$ Commented Oct 4, 2020 at 5:54
  • $\begingroup$ Right, then @AndrewChin explained the reason nicely. $\endgroup$
    – cosmo5
    Commented Oct 4, 2020 at 5:56

1 Answer 1

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In total, there are 16 balls and you need the probability of picking 3 white balls. You can consider 3 different cases each time you pick a ball, and then multiply the probabilities to get the desired answer.

Pick 1:
Total balls: 16
White balls: 5
Probability of picking white P(w): $5\over16$

Pick 2:
Total: 15
White: 4
P(w): $4\over15$

Pick 2:
Total: 14
White: 3
P(w): $3\over14$

Now that you have got the individual probabilities, the total probability of drawing 3 white balls is: $P(3W)=P(\text{W in Pick 1})\times P(\text{W in Pick 2})\times P(\text{W in Pick 3})=\Large\frac{5}{16}\times \frac{4}{15}\times \frac{3}{14}=\frac{1}{56}$

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