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jenyasd209 [6]
3 years ago
5

Set A contains three different positive odd integers and two different positive even integers; set B contains two different posi

tive odd integers and three different positive even integers. If one integer from set A and one integer from set B are chosen at random, what is the probability that the product of the chosen integers is even?
Mathematics
1 answer:
Lostsunrise [7]3 years ago
3 0

Answer:

\frac{2}{5}*\frac{3}{5} +\frac{2}{5}*\frac{2}{5}+\frac{3}{5} *\frac{3}{5}  = \frac{19}{25}

Step-by-step explanation:

We must remember that in order to get one even number we need to multiply one even number times one odd number or two even numbers. So, the first term tells the probability of having an even number from A and an even number from B, the next would be even from A and odd from B and the last one tells the likelihood of having odd from A and even from B

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A random draw is being designed for 210 participants. A single winner is to be chosen, and all the participants must have an equ
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Step-by-step explanation:  Given that a single winner is to be chosen in a random draw designed for 210 participants. Also, there is an equal probability of winning for each participant.

We are using 10 balls, numbered through 0 to 9. We are to find the number of balls which needs to be picked up, regardless of order, so that each of the 210 participants can be assigned a unique set of numbers.

Let 'r' represents the number of balls to be picked up.

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The value of 'r' can be any one of 0, 1, 2, . . , 10.

Now,

if r = 1, then

^{10}C_1=\dfrac{10!}{1!(10-1)!}=\dfrac{10!}{1!9!}=\dfrac{10\times 9!}{1\times 9!}=10

If r = 2, then

^{10}C_2=\dfrac{10!}{2!(10-2)!}=\dfrac{10!}{2!8!}=\dfrac{10\times 9\times 8!}{2\times 1\times 8!}=45

If r = 3, then

^{10}C_3=\dfrac{10!}{3!(10-3)!}=\dfrac{10!}{3!7!}=\dfrac{10\times 9\times 8\times 7!}{3\times 2\times 1\times 7!}=120

If r = 4, then

^{10}C_4=\dfrac{10!}{4!(10-4)!}=\dfrac{10!}{4!6!}=\dfrac{10\times 9\times 8\times\times 7\times 6!}{4\times 3\times 2\times 1\times 6!}=210.

Therefore, we need to pick 4 balls so that each participant can be assigned a unique set of numbers.

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Hello Guys, I really need help with these questions. Here they are.
Slav-nsk [51]

Answer:

See below.

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(b)  |x + 3| < 9 means that the distance from x + 3 to 0 is less than 9.  That puts x + 3 between -9 and 9:

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-12 < x < 6

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