Let ‘p’ be the probability of success and ‘q’ be the probability of failure.
We are given p = 2q
we know p + q = 1
On solving, we get p=

2

3

and q=

1

3

.
X has a binomial distribution with n=6,p=

2

3

,q=

1

3

∴P(X=r)=nCr(q)n−rpr
⇒ Required probability = P(at least 4 successes in next 6 trials)
= P(X ≥ 4) = P(X = 4) + P(X = 5) + P(X = 6)
=6C4q2p4+6C5q1p5+6C6q0p6
=15(

1

3

)2(

2

3

)4 +6(

1

3

)(

2

3

)5 +1×1×(

2

3

)6
=

31

9

(

2

3

)4