odds for one event happening: 1/118 odds for second event happening: 1/76 what is odds of either event happeni

Question

odds for one event happening: 1/118

odds for second event happening: 1/76

what is odds of either event happening

FIRST ANSWER 10 POINTS (correct answer)

one engineering and one computer science and we're too lazy to figure this out

NO BOTH EVENTS CANNOT HAPPEN AT THE SAME TIME!

Answer 1

Odds of first event *not* happening: 117/118

Odds of second event *not* happening: 75/76

Odds of neither event happening: (117*75)/(118*76)

Odds of at least one event happening: 1 - (117*75)/(118*76)

= .0215

Answer 2

P(either occuring) = P(event 1 happening) + P(event 2 happening) - P(both happening)

Since you said that both events can't happen P(both happening is 0 so we're just left with the sum of the two events happening.

Answer = 1/118 + 1/76 = (118+76)/(118*76) = 194/8968 = 97/4484

Answer 3

odds for event 1 happening and not event 2:
1/118 x 75/76 = a

odds for event 2 happening and not event 1:
117/118 x 1/76 = b

odds for either event taking place = a + b

Answer 4

Since both events cannot happen at the same time, it is the sum of 1/118 and 1/76: 97/4484 or 0.021632471

Answer 5

It's the sum of the two probabilities.

1/118 + 1/76 = 97/4484


Doug

Answer 6

vikram b is right. You cant in simple terms upload them considering could be counting the opportunity of having the two outcomes two times. regardless of if, vikrams coin occasion is undesirable, by using fact the threat of having a head or a tail after flipping a coin two times somewhat is one hundred% (the coin might desire to be one or the different so the possibilities of the two is a million/a million.) in certainty, the ambiguity lies in what share activities you're conversing approximately. regardless of if that's only one experience, then including them together could be superb suited. (IE if the possibilities represented areas on a extensive spinning wheel that replaced into spun as quickly as.) . regardless of if that's 2 seperate activities, (2 wheels or 2 spins of one wheel) then vikram B's approach is optimal as I clarify interior the occasion under. based o nthe wording of the question i could lean in direction of Vikrams answer. Say you have a spinner with equivalent areas for A B and C. Eahc could have a threat of a million/3. Therfore the threat of having an A or a B with one spin is a million/3 +a million/3 = 2/3. regardless of if, if ther are 2 seperate spins, the possibilities are actually not 2/3. The threat is a million/3+a million/3-a million/9. Why, by using fact there are 2 a thank you to get A out of 6 options and 2 a thank you to get B out of six options, yet by using fact you will get A and B together, including them up at as quickly as could be counting two times considered one of his options two times.

Answer 7

the odds of you posting this question rule out your theory


3 times is a bit much

Answer 8

p1/1-p1=1/118
118p1=1-p1
p1=1/119
and p2=1/77