Text Solution
Answer : ` (i) 5/12 (ii) 13/18`
Solution : (i) Number of all possible outcomes is 36. <br> Let E be the event of getting the sum less than 7 on the two dice. <br> Then, the favourable outcomes are <br> (1,1),(1,2),(1,3),(1,4),(1,5),(2,1),(2,2),(2,3),(2,4),(3,1),(3,2),(3,3),(4,1),(4,2),(5,1). <br> Number of favourable outcomes = 15. <br> ` :. P(E ) = 15/36 = 5/12`.
Distribute the referral code to your friends and ask them to register with Tutorix using this referral code.
Once we get 15 subscriptions with your referral code, we will activate your 1 year subscription absolutely free.
Your subscribed friend will also get 1 month subscription absolutely free.
The outcomes when two dice are thrown together are(1,1), (1,2), (1,3), (1,4), (1,5), (1,6)(2,1), (2,2), (2,3), (2,4), (2,5), (2,6)(3,1), (3,2), (3,3), (3,4), (3,5), (3,6)(4,1), (4,2), (4,3), (4,4), (4,5), (4,6)(5,1), (5,2), (5,3), (5,4), (5,5), (5,6)(6,1), (6,2), (6,3), (6,4), (6,5), (6,6)
Total number of outcomes = 36
(i) Let A be the event of getting the numbers whose sum is less than 7.
The outcomes in favour of event A are (1, 1), (1,2), (1,3), (1,4), (1,5), (2,1), (2,2), (2,3), (2,4), (3,1), (3,2), (3,3), (4,1), (4,2) and (5,1).Number of favourable outcomes = 15
(ii) Let B be the event of getting the numbers whose product is less than 16.
The outcomes in favour of event B are (1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (2,1), (2,2), (2,3), (2,4), (2,5), (2,6), (3,1), (3,2), (3,3), (3,4), (3,5), (4,1), (4,2), (4,3), (5,1), (5,2), (5,3), (6,1) and (6,2).
Number of favourable outcomes = 25
iii) Let C be the event of getting the numbers which are doublets of odd numbers.
The outcomes in favour of event C are (1,1), (3,3) and (5,5).
Number of favourable outcomes = 3
The outcomes when two different dice are thrown together are
(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6)
(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6)
(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6)
(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6)
(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)
(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)
∴ Total number of outcomes = 36
Let A be the event of getting the product of the numbers less than 18.
The outcomes in favour of the event A are (1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (2,1), (2,2), (2,3), (2,4), (2,5), (2,6), (3,1), (3,2), (3,3), (3,4), (3,5), (4,1), (4,2), (4,3), (4,4), (5,1), (5,2), (5,3), (6,1) and (6,2).
Number of favourable outcomes = 26
∴ PA=Favourable number of outcomesTotal number of outcomes=2636=1318
Thus, the probability that the product of the numbers appeared is less than 18 is 1318.
SOLUTION:
The outcomes when two dice are thrown together are
(1,1), (1,2), (1,3), (1,4), (1,5), (1,6)
(2,1), (2,2), (2,3), (2,4), (2,5), (2,6)
(3,1), (3,2), (3,3), (3,4), (3,5), (3,6)
(4,1), (4,2), (4,3), (4,4), (4,5), (4,6)
(5,1), (5,2), (5,3), (5,4), (5,5), (5,6)
(6,1), (6,2), (6,3), (6,4), (6,5), (6,6)
Total number of outcomes = 36 (i) Let A be the event of getting the numbers whose sum is less than 7.
The outcomes in favour of event A are (1, 1), (1,2), (1,3), (1,4), (1,5), (2,1), (2,2), (2,3), (2,4), (3,1), (3,2), (3,3), (4,1), (4,2) and (5,1).
Number of favourable outcomes = 15
∴ P(A ) = Number of favourable outcomesTotal number of outcomes=1536=512
(ii) Let B be the event of getting the numbers whose product is less than 16. The outcomes in favour of event B are (1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (2,1), (2,2), (2,3), (2,4), (2,5), (2,6), (3,1), (3,2), (3,3), (3,4), (3,5), (4,1), (4,2), (4,3), (5,1), (5,2), (5,3), (6,1) and (6,2).
Number of favourable outcomes = 25
∴ P(B ) = Number of favourable outcomes/Total number of outcomes=2536 (iii) Let C be the event of getting the numbers which are doublets of odd numbers. The outcomes in favour of event C are (1,1), (3,3) and (5,5). Number of favourable outcomes = 3
∴ P(C ) = Number of favourable outcomes/Total number of outcomes=336=112