Quants – Third Test at Oval
Q1. If it rains thrice in a week at oval what is the probability that today’s match ‘ll be affected by rain.
Q2. In last 3 weeks rain had followed a pattern as given below
18ml, 100ml, 294ml.
How much rain is going to fall during the course of the match (a test match is played for 5 days) if the trend continues?
Q3. Indian cricket team has 5 bowlers, 7 batsman and 2 wicket-keeper. In how many ways the captain can select a team of 11 with at least 4 bowlers and 1 wicket keeper.
Q4. Removed (Hint : Probability Question).
Q5. All the available 40,000 tickets are sold. If 3 out of every 4, ticket holders are fan of Dravid and 2 out of every 3, ticket holders are fan of Sachin then what is maximum & minimum number of ticket holders who are fan of both Dravid and Sachin.
n/a
Q1) is 1/7 --- I think its wrong . . .we cant ignore the statement that it rains thrice in a week.
the answer shd be 3/7
Q3. Indian cricket team has 5 bowlers, 7 batsman and 2 wicket-keeper. In how many ways the captain can select a team of 11 with at least 4 bowlers and 1 wicket keeper.
Ans: 4BOL + 1W + 6 B : Number of ways = 70
4BOL + 2W + 5B : Number of ways = 105
5BOL + 1W + 5B : Number of ways = 42
5BOL + 2W + 4B : Number of ways = 35
Total = 252
Q2. 18ml, 100ml, 294ml
2 x 32 , 4 x 52 , 6 x 72
Next number in the series is 8 x 92 = 648ml in 7 Days
So in 5 days => 463ml (approx.)
But luckily it’s not raining and 
Sachin Fan: 30,000
Dravid Fan :26,667
Maximum number of ticket holders who are fan of both sachin and dravid = 26,668
Minimum number of ticket holders who are fan of both sachin and dravid = 16,667
hi anita,
I'VE NOT IGNORED THE STATEMENT !!! I'll explain u my approach...
Probability that it rains in a week = 3/7
Probability that match is played on a rainy day= 1/3 (since it rains thrice a week)
So, probability that today's match will be affected by rain = 3/7 X 1/3 = 1/7
Did u get my pt.????
pbty of any day being a rainy day is 3/7
n the match may b on any day hence 3/7
n/a
the trend is 2*1(3^2) .....2*2(5^2).....2*3(7^2)..so nxt term is 2*4(9^2)=648
so ans is 648*5/7
n/a
hi secondharsa
how did u do all these? specially how did u calculate d minimum no of ticket holders? plz elaborate.
3/4 th of 40,000 is 30,000=sachin fans
2/3rd of 40,000 is 26,667=dravid fans
so max common is 26,667 the case where all dravid's fans are sachins fans
and minimum common would b (30,000+26,667)-40,0000=16,667
n/a
Ans to Q1) is 1/7 & ans to Q3) is 252.....tell me if i am wrong !!!!