questionr
Early access
Exams
Joint Entrance Examination
Mathematics
Probability
Question 360
Question 360 of 417
Medium
MCQ
Let in a Binomial distribution, consisting of 5 independent trials, probabilities of exactly 1 and 2 successes be 0.4096 and 0.2048 respectively. Then the probability of getting exactly 3 successes is equal to :
A
\[{{40} \over {243}}\]
B
\[{{128} \over {625}}\]
C
\[{{80} \over {243}}\]
D
\[{{32} \over {625}}\]
Check answer
Press A-D to select · Enter to check
Solutions · 1
Post your solution
Official solution
5
C
1
p
1
q
4
{}^5{C_1}{p^1}{q^4}
5
C
1
p
1
q
4
= 0.4096 ..... (1)
5
C
2
p
2
q
3
{}^5{C_2}{p^2}{q^3}
5
C
2
p
2
q
3
= 0.2048 ..... (2) \[{{(1)} \over {(2)}} \Rightarrow {q \over {2p}} = 2 \Rightarrow q = 4p\] \[p + q = 1 \Rightarrow P = {1 \over 5},q = {4 \over 5}\] P (exactly 3) = \[{}^5{C_3}{(p)^3}{(q)^2} = {}^5{C_3}{\left( {{1 \over 5}} \right)^3}{\left( {{4 \over 5}} \right)^2}\] \[ = 10 \times {1 \over {125}} \times {{16} \over {25}} = {{32} \over {625}}\]
Answer the question to unlock
Previous
Next