If X and Y are independent Binomial variates B(5,1/2) and B(7,1/2), then P(X+Y=3) is equal to:
3547
551024
220512
11204
Explanation for the correct option.
Step 1: Find the value of PX=0,Y=3
We know that, P(X+Y=k)=∑a,bCanpn·Cbmqm
PX=0,Y=3=C05125×C37127=1×132×7×6×5×4!3×2×1×4!×1128=351096.........(1)
Step 2: Find the value of PX=1,Y=2
PX=1,Y=2=C15125×C27127=5×132×7×6×5!2×1×5!×1128=1051096.........(2)
Step 3: Find the value of PX=2,Y=1
PX=2,Y=1=C25125×C17127=5×4×3!2×1×3!×132×7×1128=701096.........(3)
Step 4: Find the value of PX=3,Y=0
PX=3,Y=0=C35125×C07127=5×4×3×2!3×2×1×2!×132×1×1128=101096.........(4)
Step 5: Find the value of P(X+Y=3)
By adding 1,2,3,4 we get,
P(X+Y=3)=354096+1054096+704096+104096=2204096=551024
Hence, option B is correct.