36
36
Q.11. Two friends are travelling in a train. As they were feeling bored,
they started playing a game with a pair of dice. Each one of
them started rolling the pair of dice one by one stating one.
condition before rolling. If one person gets the numbers
according to the condition stated decided by him, he wins and get a score. Now answer
any four of the following-
(i) Second friend says, "sum is an even number". The probability of his losing is-
(a) less than
(b) more than
(c) equal to
(d) 1
(ii) First friend says, 6 will come up either time.'. the probability of his loosing is -
25
(a)
(b)
(c);
(d)
(iii) Second friend says, 'sum less than 9'. The probability of his winning is -
(a)
(b)
(c)
(d)
(iv) First friend says "a doublet" The probability of his winning is
(a)
(b)
(c)
(d)
(v) The probability of getting sum 15 is -
(a) 1
(b) 0
(c)
(vi) No. of possible outcomes on a rolling pair of dice is -
(a) 6
(b) 12
(c) 8
(d) 36
31
5
(d)
Answers
i) The probability of losing if the sum is an even number is equal to the probability of winning if the sum is an odd number.
ii) The probability of losing if 6 will come up either time is 0.66
iii) The probability of winning if the sum will be less than 9 is 0.72
iv) The probability of winning with a doublet is 0.16
vi) The number of outcomes in a pair of dice is going to be 36
vii) The probability of getting a sum of 15 is 0 as it is an unlikely event.
Given:-
There are two dice.
∴There are 36 possible outcomes (refer to the table attached below)
To Find:-
i) Probability of losing if the sum is an even number
ii) Probability of losing if 6 comes up either time
iii) Probability of winning if the sum is less than 9
iv) Probability of winning if it's a doublet
v) Probability of getting a sum of 15
vi) No. of outcomes on a pair of dice
Solution:-
i) The probability of losing if the sum is an even number is equal to the probability of winning if the sum is an odd number. There are 18 possible even outcomes and 18 possible odd outcomes. (Refer to the table) Therefore using the probability formula;
P(even number)=n/N
P(even number)=18/36
P(even number)=1/2
Since we care considering losing we need to subtract the probability of getting an even number with 1 which is 1/2
ii) There are 12 possible outcomes where 6 will come up either time.
Therefore,
P(losing if 6 comes up either time) = 1 - (n/N)
P(losing if 6 comes up either time) = 1 - (12/36)
P(losing if 6 comes up either time) = 1 - (1/3)
P(losing if 6 comes up either time) = 2/3
∴The probability of losing if 6 will come up either time is 2/3 or 0.66
iii) There are 26 possible outcomes where the sum will be less than 9.
Therefore,
P(winning if sum will be less than 9) = (n/N)
P(winning if sum will be less than 9) = (26/36)
P(winning if sum will be less than 9) = (13/18)
P(winning if sum will be less than 9) = 0.72
∴The probability of winning if the sum will be less than 9 is 0.72
iv) There are 6 possible outcomes where there is a doublet
Therefore,
P(winning with a doublet) = (n/N)
P(winning with a doublet) = (6/36)
P(winning with a doublet) = (1/6)
P(winning with a doublet) = 0.16
∴The probability of winning with a doublet is 0.16
v) The probability of getting a sum of 15 is 0 as it is an unlikely event. Since we have two dice that have six sides the highest sum we can get is 12 i.e. (6+6), therefore we will not get anything higher than 12.
vi) The number of outcomes in a pair of dice is going to be 36 as there are 6 outcomes in one die and 6 in the other. Since both of them are tossed together their outcomes become 36.
#SPJ1
I If the sum is an even number, the odds of losing are the same as the odds of winning if the sum is an odd number.
ii) If the number 6 appears both times, there is a 0.66 % chance of losing.
iii) If the sum is less than 9, there is a 0.72 % chance of winning;
iv) there is a 0.16 % chance of winning with a doublet; and
v) there will be 36 outcomes from a pair of dice.
vi) Since getting a total of 15 is improbable, there is no chance of doing so.
Given:
There are two dice available.
There are 36 potential outcomes
To Find:
i)Chance of losing if the total is an even number.
ii) Chance of losing if 6 appears both times.
iii) Chance of winning when the total is less than 9
iv) Chances of winning if the wager is a doublet
v) Chances of receiving a 15-dollar amount
vi) The number of results from a pair of dice
Solution:-
i) If the sum is an even number, the odds of losing are the same as the odds of winning if the sum is an odd number. There are 18 even outcomes and 18 odd outcomes that could occur. Applying the probability formula as a result;
P= for even numbers
P= for even numbers
P= for even numbers
We must remove the likelihood of receiving an even number with 1, which is , because we care about losing.
ii) Of the 12 possible outcomes, 6 will occur in every case.
Therefore,
P(falling short if 6 appears any way) = 1 - (n/N)
P(falling short if 6 appears any way) = 1 - (12/36)
P(falling short if 6 appears either way) = 1 - (1/3)
P(falling short if 6 appears either way) = 2/3
If the number 6 appears both times, the likelihood of losing is 2/3, or 0.66.
iii) There are 26 scenarios in which the total will be fewer than nine.
Therefore,
P(winning if total is less than 9) = (n/N).
P(winning if total is less than 9) = (26/36)
P(winning if total is less than 9) = (13/18)
P(winning if the total is fewer than nine) = 0.72
If the total is fewer than 9, there is a 0.72 percent chance of winning.
iv) When there is a doublet, there are six alternative outcomes.
Therefore,
P(achieving a doublet victory) = (n/N)
P(taking a doublet and winning) = (6/36)
P=1/6 for winning with a doublet.
P(getting a doublet win) = 0.16
With a doublet, the likelihood of winning is 0.16.
v) Since getting a total of 15 is improbable, there is no chance of doing so. The largest amount we can get with two dice that each have six sides is 12, or (6+6), therefore we won't get anything else.
vi) A pair of dice will have 36 possible results since each die has 6 possible outcomes and each die has 6 possible outcomes.