They are more like the conditional probability, first we have to figure out what it is the condition is and what would be the formality of the bayesian rule.
(1)
“You’re about to get on a plane to Seattle. You want to know if you should bring an umbrella.
“You call 3 random friends of yours who live there and ask each independently if it’s raining.
“Each of your friends has a 2/3 chance of telling you the truth and a 1/3 chance of messing with you by lying.
“All 3 friends tell you that ‘Yes’ it is raining. What is the probability that it’s actually raining in Seattle?”
The prior probability of raining is 40%., the probability of not raining is 60%.
P(raining | yes) = [P( yes | raining) * P(raining )] / P(YES)
since P (Yes) could be calculated as:
P(YES) = P (yes|raining)P(raining) + P(yes|not raining)P(not raining)
P(YES) = (0.6)^30.4 + (0.4)^30.6 = 0.0864 + 0.0384= .1248
Therefore, the
P(raining | yes) =0.0864/ .1248 = 0.6923 ,so bring the umbrella.
(2)
“Pick up a coin C1 given C1+C2 with probability of trials p(h1)=.7, p(h2)=.6 and doing 10 trials.
“What is the probability that the given coin you picked is C1 given you have 7 heads and 3 tails?”
p(h1)=.7 p(t1)=.3
p(h2)=.6
P(C1| 7 heads,3 tails) = [P(7 heads , 3 tails | C1 )*P(C1)/P(7heads, 3tails) ]
P(7heads, 3tails)=P(7 heads , 3 tails | C1 )P(C1)+P(7 heads , 3 tails | C2 )P(C2)
Therefore
P(7heads, 3tails) = C(10, 7) (0.7)^7 (0.3)^30.5 + C(10,7)(0.6)7*(0.4)30.5
= C(10,7) 0.5 (A +B)
= 60 0.04 =0.24
P(7 heads , 3 tails | C1 )P(C1)= 0.5120* 0.0022 = 0.132
Therefore: P(C1| 7 heads,3 tails) = 0.132 /0.24 = 0.55
(3)
Consider a game with 2 players, A and B. Player A has 8 stones, player B has 6. Game proceeds as follows.
First, A rolls a fair 6-sided die, and the number on the die determines how many stones A takes over from B.
Next, B rolls the same die, and the exact same thing happens in reverse.
This concludes the round. Whoever has more stones at the end of the round wins and the game is over. If players end up with equal # of stones at the end of the round, it is a tie and another round ensues. What is the probability that B wins in 1, 2, ..., n rounds?
Extreme cases:
Key is to define the problem correctly. Let Na be the # player A rolls, and Nb be the number player B rolls. Then at the end of the first round A will have (8 + Na - Nb) and B will have (6 - Na + Nb) stones. So all you need is to compute Pr{6 - Na + Nb > 8 + Na - Nb} for round 1 victory. Subsequent rounds are even easier since all subsequent rounds can only start when the stone count is equal.
(4)
How many birthday posts occur on Facebook on a given day?
Total users in Facebook : 1.2 billion
equal distribution of each day in birthday: 365 / Leap year
facebook posts percent: 70%
(1.2billion /365) * 0.7 = 2301369 =2.3 million posts every day
