Test: Probability (CBSE Level) - 1
25 Questions MCQ Test Mathematics For JEE | Test: Probability (CBSE Level) - 1
Probability of getting a number between 1 and 100, which is divisible by 1 and itself only, is
Prime numbers between 1 and 100: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97 n (Prime numbers) = 25
n (Total numbers between 1 and 100) = 98
P (getting a prime number) = 25/98
Two players toss four coins each. The probability that both obtain the same number of heads is
The tossing of coin by each of the two players is independent. So we can multiply the respective probabilities and get the required probability.
There are five cases:
Zero Heads
The probability that A(Let us call the players A&B) gets no head is 4C0 × (½)4 = 1/16
The probability that both of them get zero heads is 1/16×1/16 = 1/256
One Head
The probability that A gets 1 head is 4C1×1/2×(½)3 = 4×1/16 = 1/4
The probability that both A & B both get 1 head=1/4×1/4
= 1/16
Two Heads
Calculating in a similar manner,the proability that A gets two heads is,
4C2 × (½)2 × (½)2
= 3/8
The probability that both get two heads = 9/64
Three Heads
The probability that A gets 3 heads is,
4C3× (1/2)3 ×1/2
= 1/4
The probability that both get three heads = 1/16
Four Heads
The probability that A gets 4 heads is, 4C4×(½)4
= 1/16
The probability that both get four heads=1/256
So,required probability = 1/256 + 1/16 + 9/64 + 1/16 + 1/256
= 70/256
= 35/128
If three dice are thrown, then the probability that they show the numbers in A.P. is
A.P. with common diff one are =4=(123)(234)(345)(456)
A.P. with common diff 2 are =2=(135)(246)
A.P. with common diff 0 are =6
⇒P=2+4+6/216=1/18
A cubical dice has 3 on three faces, 2 on two faces and 1 on the 6 thth face .It is tossed twice. The chance that both the tosses show an even number is
The correct option is B.
Since we have only even number 2,so when the dice is rolled twice we have (2,2) four times which means 2 on first dice pairs with both the twos on the second dice , similarly second two on first dice pairs with both the twos on the second dice
Hence the probability = 4/36 = 1/9
A coin is tossed once. If a head comes up, then it is tossed again and if a tail comes up, a dice is thrown. The number of points in the sample space of experiment is
A drawer contains 5 black socks and 4 blue socks well mixed. A person searches the drawer and pulls out 2 socks at random. The probability that they match is
Out of 9 socks, 2 can be drawn in 9C2 ways.
Therefore, the total number of cases is 9C2.
Two socks drawn from the drawer will match if either both are black or both are blue.
Therefore, favorable number of cases is 5C2+ 4C2.
Hence, the required probability is
(5C2+ 4C2.)/ 9C2
= 4/9
A bag contains 5 white, 7 red and 4 black balls. Four balls are drawn one by one with replacement. The chance that atleast two balls are black is
A and B take turn in throwing a pair of dice. A wins if he throws a total of 5 before B throws a total of 7. If A has the first throw, the probability of his winning the game is
A determinant is chosen at random from the set of all determinants of order 2 with elements 0 or 1 only. The probability that the value of the determinant chosen is positive is
If the letters of the word ‘ INDEPENDENCE ‘ are written down at random in a row, then the chance that no two E’s occur together is
The letters of the word ‘ASSASSIN ‘ are written at random in a row. The chance that all the similar letters occur together is
Total number of ways to arrange "ASSASSIN" is 8!/(4!.2!) first we fix the position.
A person throws successively with a pair of dice. The chance that he throws 9 before he throws 7 is
Let A be set of 4 elements. From the set of all functions from A to A, a function is chosen at random. The chance that the selected function is an onto function is
Both A and B throw a dice. The chance that B throws a number higher than that thrown by A is
Both A and B throw a dice. The chance that B throws a number not less than that thrown by A is
Two dice are thrown. The number of sample points in the sample space when 6 does not appear on either dice is
8 coins are tossed at a time. The probability of getting 6 heads up is
he required probability = 
= 
A dice is rolled 6 times. The probability of obtaining 2 and 4 exactly three times each is
A dice has 3 faces each bearing ‘ 2 ‘ and three faces each bearing ‘ 6 ‘. It is rolled once. The probability of showing up ‘a six ‘ is
S = {2, 2, 2, 6, 6, 6}
n(S) = 6
n(6) = 3
P(showing up a 6) = n(6)/n(S) = 3/6 = 1/2
The total area under the standard normal curve is
Neelam and Nidhi throw with a dice. The chance that both Neelam and Nidhi throw the same number is
A dice is tossed once and even number has come up. The chance that it is either 2 or 4 is
What is the probability of getting a sum of eight if two dice are thrown at once?
Explanation : (1,1) (1,2) (1,3) (1,4) (1,5) (1,6) (2,1) (2,2) (2,3) (2,4)
so on upto (6,6) are the total numbers we get when we throw a pair of dice that is, 36
In the question they are asking the probability of getting a sum of 8
(2,6) (3,5) (4,4) (5,3) (6,2) are the pairs which upon adding we get 8
Therefore, 5/36 is your answer
The probability of having atleast one head in 5 throws of a coin is
A fair coin is tossed a fixed number of times. If the probability of getting 4 heads is equal to the probability of getting 7 heads, then the probability of getting 2 heads is
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