Multiple choice

A bag contains three tickets numbered $1, 2$ and $3$. A ticket is drawn at random and put back in the bag, and this is done four times. The probability that the sum of the numbers drawn is even is

  1. $\dfrac {40}{81}$
  2. $\dfrac {41}{81}$
  3. $\dfrac {14}{27}$
  4. $\dfrac {13}{81}$
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B Correct answer
Explanation

Let p be the probability of drawing an even number (2) in one trial, p = 1/3. Let q be the probability of drawing an odd number (1 or 3), q = 2/3. The sum is even if we draw an even number of odd-numbered tickets (0, 2, or 4). Using the binomial expansion (p+q)^4 and (p-q)^4, the sum of even terms is 1/2 * ((p+q)^4 + (p-q)^4). Here p=1/3, q=2/3, so 1/2 * (1^4 + (-1/3)^4) = 1/2 * (1 + 1/81) = 1/2 * (82/81) = 41/81.

AI explanation

For the sum of the numbers drawn to be even over four draws, there must be an even number of odd tickets drawn, meaning either 0, 2, or 4 odd tickets are selected. The probability of drawing an odd number is 2/3 and an even number is 1/3. Using the binomial probability formula for these three cases gives (1/3)^4 + 6*(2/3)^2*(1/3)^2 + (2/3)^4, which evaluates to 1/81 + 24/81 + 16/81. Adding these probabilities together gives a total probability of 41/81.

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