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January 10th, 2018, 09:27 AM   #1
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Need help to solve below problems?

1)
If a fair coin (a coin which shows heads or tails with equal probabilities when tossed) shows heads upon tossing, 4 numbers are selected from 1, 2, 3, ... 100 (a number once selected is not selected again) and if tails shows up, 3 numbers are drawn from 3, 6, 9, 12, ... 99 (a number once selected is not selected again). Determine the probability of the event that the sum of the selected numbers is even.

2)
Iron deficiency is known to exist in 28%, 14% and 5% of the people of three different tribes T1, T2 and T3 respectively. All the three tribes have equal population. An investigator selects a person from one of the tribes randomly, from the total population of the three tribes and examines for iron deficiency. It turns out that the selected person does not have the deficiency. Determine the probability that this person belonged to tribe (i) T1, (ii) T2, (iii) T3.
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January 10th, 2018, 09:52 AM   #2
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With respect to three numbers, if all three are odd, the sum is odd. If only two are odd, the sum is even. if only one is odd, the sum is even. And if none is odd, the sum is even.

So GIVEN that three numbers are selected, what are the respective probabilities that all three are odd and that only one is odd?

So GIVEN that three numbers are selected, what is the probability that their sum is odd?

Now do the same analysis for four numbers.

GIVEN that four numbers are selected, what is the probability that their sum is odd?

Now what?
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January 10th, 2018, 12:19 PM   #3
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Quote:
Originally Posted by JeffM1 View Post
. . . if only one is odd, the sum is even.
???
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January 10th, 2018, 03:21 PM   #4
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Quote:
Originally Posted by skipjack View Post
???
Oops.

Basically a typo, but still!!!
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