There are 12 card numbered 1 to 12 in a box. If tow cards are selected, what is the probability that the sum is odd (I) with replacement (II) without replacement.
SOLUTIONS (1/4,6/11)
HOW I CAN SOLVE WITH REPLACEMENT.
Knowing that
Sample space (S) = 12C2=66
And there are 6 odd (O)& 6 even (E)
So, probability without replacement= (6C1*6c1)/66 =6/11
So how to solve this problem with replacement. Thanks
To solve this problem with replacement, you need to consider the probabilities of selecting an odd card and an even card separately.
With replacement means that after each draw, you put the card back into the box. This means that the sample space remains the same for both draws, which is 12 cards.
To find the probability of selecting an odd card, you need to determine the number of odd cards in the box. Since there are 6 odd cards (O) out of the total 12 cards, the probability of selecting an odd card on each draw is 6/12 or 1/2.
Now, to find the probability of selecting an even card, you subtract the probability of selecting an odd card from 1 (since the sum of the probabilities must equal 1). So, the probability of selecting an even card on each draw is 1 - 1/2 = 1/2.
To find the probability of the sum being odd, you need to consider the possible combinations of odd and even cards. There are two scenarios to consider:
1. Odd card + Odd card
2. Even card + Even card
For the first scenario, the probability is (1/2) * (1/2) = 1/4, as you are selecting an odd card on both draws.
For the second scenario, the probability is (1/2) * (1/2) = 1/4, as you are selecting an even card on both draws.
Since these two scenarios are mutually exclusive (they cannot occur simultaneously), to find the overall probability, you add the probabilities of individual scenarios.
Therefore, the probability of the sum being odd with replacement is 1/4 (from scenario 1) + 1/4 (from scenario 2) = 2/4 = 1/2.
In summary, the probability of selecting two cards with a sum that is odd with replacement is 1/2.