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Six billiard balls, numbered $1$ through $6,$ are placed in a box. Three of the balls are red, and three are blue. One ball is to be drawn randomly from the box.

i. The probability that the ball drawn will be an even numbered red ball

ii. $\large \frac{1}{2}$

Question: Is Option i. greater than Option ii?


My attempt:

Probability of drawing $1$ red ball = $\large \frac{1}{2}.$

Probability of drawing a red ball AND a even number is $\large \frac{1}{2}\cdot\frac{1}{2}=\frac{1}{4}.$

Therefore my answer is no, since $\large \frac{1}{4}<\frac{1}{2}.$

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    $\begingroup$ Sounds good reasoning to me. $\endgroup$ – EA304GT Nov 12 '15 at 1:39
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    $\begingroup$ Yes, although, that is assuming that there's an unbiased probability that any even ball is red. But in the worst case scenario where all even balls are red, then the probability of drawing an even and red ball is $1/2$. $\endgroup$ – Graham Kemp Nov 12 '15 at 2:10
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The probability that you draw an even numbered ball is ${1\over2}$. From $P(A\cap B)\leq P(A)$ it follows that the compound probability in question cannot be $>{1\over2}$ whatever the mechanism in choosing the colors of the balls had been.

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