You pick a card at random. Without putting the first card back, you pick a second card at random 3,4,5,6 .What is the probability of picking an even number and then picking and odd number ?

3/10

Step-by-step explanation:

I'm going to assume you are starting with five cards, since if you pulled one and that left four cards, there must have been five cards at the start.

If you had five cards 2, 3, 4, 5, and 6, the probability of taking an even number would be 3/5 since there are three even numbers out of a total of five cards. Then, from there, you would need to find the probability of picking an odd number from the four cards left: 3, 4, 5, and 6. The probability of picking an odd numbered card from that would be 2/4. We would then need to simplify that which gives us 1/2. From here, we need to multiply 3/5 by 1/2 in order to get the probability.

3/5 x 1/2 = 3/10

-14 is the answer

hope this : ))

since proposition states that a number is greater than 25, if is true then the number must actually be greater than 25.

this implies that all possible options, except 24, can be accepted