Two dice are thrown simultaneously. What is the probability of getting two numbers whose product is even?
In a simultaneous throw of two dice, we have n(S) = (6 x 6) = 36.
| Then, E | = {(1, 2), (1, 4), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 2), (3, 4), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 2), (5, 4), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)} |
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n (E) = 27.
If 4 numbers at random are multiplied together, the probability that the last digit in the product is 1,3,7 or 9 is
If the last digit in the product of four numbers has to end in 1,3,7 or 9 then each of the four numbers has to ene in 1,3,7 or 9 .
Therefore of the ten digit four are favourable and probability is 4/10 =2/5
Therefore the probablity of each of the four number ending in 1,3,7 or 9 is
(2/5)4 =16/625
The probability that Soumya will get marry with in 365 days is ‘a’ and the probability that her colleague Alma get marry within 365 days is ‘b’. Find the probability that only one of the two get marry at the end of 365 days.
The probability that Soumya will get marry than Alma will be =a(1-b)
Similarly The probability that Alma will get marry than Soumya will be
b(1-a)
The probability that either of these two get marry =a(1-b) +b(1-a) =a+b-2ab
If six persons sit around a table, the probability that some specified three of them are always together is
There are six persons and three of them are grouped together.Since it is a circle, this can be done in 3! 3! ways.
:. P =(3! 3!)/5! = 3/10
If a number is chosen at random from 1 to 100, then the probability that the chosen number is a perfect cube is
We have 1,8,27 and 64 as perfect cubes from 1 to 100.
Thus, the probability of picking a perfect cube is 4/100 = 1/25
If a coin is tossed twice what is the probability that it will land either heads both times or tails both times?
Note the “or” in between event A (heads both times) and event B (tails both times). That means more options, more choices, and a higher probability than either event A or event B individually. To figure out the probability for event A or B, consider all the possible outcomes of tossing a coin twice: HH, TT, HT, TH Since only one coin is being tossed, the order of heads and tails matters. Heads, tails and tails, heads are sequentially different and therefore distinguishable and countable events. We can see that the probability for event A is ¼ and that the probability for event B is ¼. We expect a greater probability given more options, and therefore we can eliminate choices a, b and c, since these are all less than or equal to ¼. Now we use the rule to get the exact answer. P(A or B) = P(A) + P(B). If either event 1 or event 2 can occur, the individual probabilities are added: ¼ + ¼ = 2/4 = ½.
A question has 7 options. If he chooses correct answer he can get one mark. If he chooses wrong option he loses one mark. If he chooses randomly he gets zero marks. Any way he identifies 2 options and eliminates them. If he chooses randomly how many marks can he gain?
Probability to choose correct option=1/7 and associated marks is 1;
Probability to choose incorrect option=6/7 and let the associated mark as x;
Total marks= [(1/7)*1]+[(6/7)*x]=0, i.e. x= (-1/6);
Eliminating two options,
Total marks= [(1/5)*1]+[(4/5)*(-1/6)]=1/15.
A card is drawn from a pack of 52 cards. The probability of getting a queen of club or a king of heart is:
Here, n(S) = 52.
Let E = event of getting a queen of club or a king of heart. Then, n(E) = 2.
P(E) =n(E)/ n(S)= 2/ 52= 1/ 26
What is the probability that a number selected from numbers 1,2,3,…,30, is prime number, when each of the given numbers is equally likely to be selected ?
X= {2,3,5,7,11,13,17,19,23,29}
Two people agree to meet on January 9, 2005 between 6.00 P.M. to 7.00 P.M., with the understanding that each will wait no longer than 20 minutes for the other. What is the probability that they will meet ?
They can meet when A comes between 6: 00 = 6: 40.
And so B can join him between 6: 20 = 7: 00.
Similarly, the process can be reversed.
Therefore p=(40 / 60)2 = 4/9.