## Question

Let *X* denote the number of time heads occur in *n* tosses of a fair coin. If*P*(*X* = 4), *P*(*X* = 5) and P(*X* = 6) are in *AP*; the value of *n* is:

### Solution

7

⇒ Simplify to get: *n* = 7

#### SIMILAR QUESTIONS

If two events A and B are such that

One ticket is selected at random from 100 tickets numbered 00, 01, 02,…99. Suppose A and B are the sum and product of the digits found on the ticket. Then *P*(*A* = 7/8 = 0) is given by:

A box contain 100 tickets numbered 1, 2,…100. Two tickets are chosen at random. It is given that the greater number on the two chosen tickets is not more than 10. The probability that the smaller number is 5 is:

A bag contains four tickets marked with numbers 112, 121, 211, 222. One ticket is drawn at random from the bag. Let

*E _{i}*(

*i*= 1, 2, 3) denote the event that its digit on the ticket is 2. Than:

Two dice are rolled one after the other. The probability that the number on the first is smaller than the number on the second is:

There are 20 cards. 10 of these cards have the letter ‘I’ printed on them and the other 10 have the letter ‘T’ printed on them. If three cards are drawn without replacement and kept in the same order, the probability of making word IIT is:

A box contains 2 black, 4 white and 3 red balls. One ball is drawn at random from the box and kept aside. From the remaining balls in the box, another ball is drawn at random and kept aside the first. This process is repeated till the balls are drawn from the boxes. The probability that the balls drawn are in the sequence of 2 black, 4 white and 3 red is:

A bag contains ‘*m*’ white and ‘*n*’ black balls. Two players A and B alternately draw a ball from the bag, replacing the ball each time after draw. A beings the game. If the probability of A wining (that is drawing a white ball) is twice the probability of B wining, then the ratio *m*: *n* is equal to:

If X is *a* binomial variate with parameters *n* and *p*, where 0 < *p* < 1 such that is independent of n and *r*, then *p* equals:

A bag contains 4 tickets numbered 1, 2, 3, 4 and another bag contains 6 tickets numbered 2, 4, 6, 7, 8, 9. One bag is chosen and a ticket is drawn. The probability that the ticket bears the number 4 is: