Question

A pair of coins is tossed a fixed number of times. If the probability of getting both heads exactly 3 times is same as the probability of getting both heads exactly 4 times then the number of trials is

Answer 1

ANSWER:

The number of outcomes = HH    TT    HT    TH

                                            = 4

Probability of getting

both heads 3 times     = 1 / 3

Probability of getting

both heads 4 times     = 1 / 4

$\rule{400}2$

PLZ MARK BRAINLIEST

Answer 2

Answer: Total no of times the coin is tossed is 15

Step-by-step explanation:

let the number of times the coin is tossed be n

Sample space of event = [HH,HT,TH,TT]

hence , P(getting both head) = 1/4

so, P(getting both head 3 times out of n times) = nC3 . (1/4)^3 . (3/4)^(n-3)

similarly, P(getting both head 4 times out of n times) = nC4 . (1/4)^4 (3/4)^(n-4)

A/Q. P(getting both head 3 times out of n times) = P(getting both head 4 times out of n times)

i.e.  nC3 . (1/4)^3 . (3/4)^(n-3) = nC4 . (1/4)^4 (3/4)^(n-4)

upon solving the above equation we get n=15 which is the required answer.

calculation tip : add up the powers of same terms, use the formula nCr = n!/(r! . (n-r)!). a! = a.(a-1)! remember n-3 > n-4.