Suppose that 35 people are divided in a random mannerinto two teams in such a way that one team contains10 people and the other team contains 25 people. what isthe probability that two particular people a and b will beon the same team

To be precise, the size of your sample space is (2410)(2410). This number does go on the bottom of the fraction, and what goes on top is the size of the event. Break up the event into independent events 1. choose the 2 defective bulbs, and 2. choose the remaining 8 bulbs. I don't have much choice in event 1. There is only one way to choose both of the defective balls. In other words, (22)(22) (choosing 2 defective bulbs from a set of 2 defective bulbs). For event 2, there are 24−2=2224−2=22 nondefective bulbs, and I must choose 88 of them, so that's (228)(228). Finally, since events 1 and 2 are independent, we multiply the answers for the combined event: (22)(228)(22)(228)

Or, since (22)=1(22)=1,

Hope this helps!

step-by-step explanation:

y=mx+b

the answer will be the last one