Multiple choice

$\bullet$ If $A$ and $B$ are two events associated with a random experiment, then $P(A\cup B)=P(A)+P(B)-P(A\cap B)$. $\bullet$ If $A, B, C$ are three events associated with a random experiment, then $P(A\cup B\cup C)=P(A)+P(B)-P(A\cap B)-P(B\cap C)-P(A\cap C)+P(A\cap B\cap C)$. Two car ds are drawn from a pack of $52$ cards, then the probability that either both are red or both are kings is

  1.  $\dfrac {55}{221}$.

  2. 0

  3. Cant be determined

  4. None of the above
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A Correct answer
Explanation

Total cards = 52. Red cards = 26. Kings = 4. Red kings = 2. P(Red or King) = P(Red) + P(King) - P(Red and King) = 26/52 + 4/52 - 2/52 = 28/52 = 7/13. The question asks for two cards drawn. P(both red) = (26/52)(25/51) = 25/102. P(both kings) = (4/52)(3/51) = 1/221. P(both red and both kings) = P(both red kings) = (2/52)*(1/51) = 1/1326. Using inclusion-exclusion for two cards: 25/102 + 1/221 - 1/1326 = (325 + 6 - 1) / 1326 = 330/1326 = 55/221.

AI explanation

Using the addition theorem of probability, the formula is P(A union B) equals P(A) plus P(B) minus P(A intersection B). The probability of drawing two red cards from 52 is 26C2 divided by 52C2, which equals 325 divided by 1326. The probability of drawing two kings is 4C2 divided by 52C2, which equals 1 divided by 221. The probability of drawing two red kings, which is the intersection, is 2C2 divided by 52C2, equalling 1 divided by 1326. Applying the formula gives (325 divided by 1326) plus (6 divided by 1326) minus (1 divided by 1326), which equals 330 divided by 1326, simplifying to 55 divided by 221.

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