J, can you help me wi this qn.
A new york city cab had a hit and run. Out of the 5 witnesses, 4 said the colourbof the cab was green and one said it to be yellow. The probability that the colour was rightly identified was 2/3. If 85% of the registered cabs in new york are yellow and 15% is green what is the probability that cab is green in colour
“The probability that the colour was rightly identified was 2/3.” This sentence seems to introduce some ambiguity (the interpretation I take from it seems incompatible with the other information given). Otherwise it seems to be the kind of problem one would solve by making a tree.
Three friends have their bdays in same week but on different days. What is the probability that two of them will have their bdays on weekdays and one of them has it on weekend?
Applying first principles:
Sample space = 7 x 6 x 5 = 210.
Favourable cases: 3 ways of choosing which one is on the weekend, 2 ways for that person and 5 and 4 ways for the other two = 3 x 2 x 5 x 4 = 120
So I guess it should be 120/210 = 4/7
Hello Sir, shouldn’t it be 3*2*5*4 since two friends who have their birthdays on weekdays have it on two different days, so it one out of 5 days goes to second friend it leaves only 4 other days for the third friend.
Can you please help me understand what am I missing out here?
Regards
Oops yes I missed the different days part, you’re right. Denominator will change too accordingly. Will edit the earlier response.
CAT-holics 
J, can you help me wi this qn.
A new york city cab had a hit and run. Out of the 5 witnesses, 4 said the colourbof the cab was green and one said it to be yellow. The probability that the colour was rightly identified was 2/3. If 85% of the registered cabs in new york are yellow and 15% is green what is the probability that cab is green in colour
“The probability that the colour was rightly identified was 2/3.” This sentence seems to introduce some ambiguity (the interpretation I take from it seems incompatible with the other information given). Otherwise it seems to be the kind of problem one would solve by making a tree.
regards
J
J,
Can u pls help me with the below Ques.
Three friends have their bdays in same week but on different days. What is the probability that two of them will have their bdays on weekdays and one of them has it on weekend?
Applying first principles:
Sample space = 7 x 6 x 5 = 210.
Favourable cases: 3 ways of choosing which one is on the weekend, 2 ways for that person and 5 and 4 ways for the other two = 3 x 2 x 5 x 4 = 120
So I guess it should be 120/210 = 4/7
regards
J
Hello Sir, shouldn’t it be 3*2*5*4 since two friends who have their birthdays on weekdays have it on two different days, so it one out of 5 days goes to second friend it leaves only 4 other days for the third friend.
Can you please help me understand what am I missing out here?
Regards
Oops yes I missed the different days part, you’re right. Denominator will change too accordingly. Will edit the earlier response.
regards
J