Favourable numbers are 11, 21, 31, 41.
? Required probability = 4/49
First we choose 1 machine out of the given 4 , the probability that it is faulty is 2/4 (favorable outcomes/possible outcomes)
Now we have to pick the second faulty machine the probability of doing so is 1/3.
Hence overall probability is 2/4 x 1/3 = 1/6
Total ways = 52
There are 13 cards of diamond, 4 cards of king, but one card is king of diamond which is counted both in diamond and king cards
? Favourable ways = 13 + 4 - 1 = 16
? Required probability = 16/52 = 4/13
Total ways 52C3 = 22100
There are 4 suit in a pack of cards so three suit can be selected in 4C3 ways.
One card each from different unit can selected as = 13C1 X 13C1 X 13C1 ways
So, favourable ways = 4C3 X 13C1 X 13C1 X 13C1 = 8788
? Required probability = 8788/22100 = 169/425
Total ways = 100
Squares of following no's lie between 1 and 100,
12, 22, 32 , 42, 52 , 62, 72, 82 , 92 , 102
(Which are 10 in numbers.)
So. Required probability = 10/100 = 1/10
Total number of cases is 17.
? Number divisible by 3 are 3, are 3, 6, 9, 12, 15 (These are 5 in number )
Number divisible by 7 are 7, 14. (These are 2 in number )
There are two favourable number of cases
Total no. of favourable number = 5 + 2
Required probability = 7/17.
Since, the 1st students can have any day as his birthday and according to the question corresponding to the 1st person and 2nd the 3rd person need to have the same day as their birthday and thus probability that they have identical birthday
= 1x (1/365) x (1/365)
= 1/(365)2
it is given that last 3 digits are randomly dialed
Then, each of the digit can be selected out of 10 digits in 10 ways. Hence, required probability
= 1/(10)3 = 1/1000
A non-leap year has 365 days, each day repeats for 52 times with 1 day left
This one can be {Sun, Mon, Wed, Thu, Fri, and Sat}.
? p = 6/7
Two cards can be drawn from a pack of 52 playing cards in 52C2 ways. i,e., 52 x 51 / 2 = 1326 ways. The event that two kings appear in a single drawn of cards is 4C2 ways, i.e 6 ways.
? The probability that the two cards drawn from a pack of 52 cards are kings = 6/1326 = 1/221
Since, all possible hypothesis regarding the colour of the balls are equally likely, therefore these could be 3 white balls, initially in the bag.
? Required probability = 1/4 [1 + 3/4 + 1/2 + 1/4]
= 1/4 [(4 + 3 + 2 + 1)/4] = 5/8
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