Solution
Step 1: P(wn) ki value nikalna:
Hume pata hai ∑P(wn)=1.
Ye ek GP banati hai: a+2a+4a+⋯=1⟹1−1/2a=1⟹2a=1⟹a=21.
Isliye, P(wn)=2n1.
Step 2: Set A ke elements dhundna:
A={2k+3l:k,l∈N}. Sabse choti value jab k=1,l=1 hai: 2(1)+3(1)=5.
k=2,l=1⟹7; k=1,l=2⟹8; k=3,l=1⟹9.
Dhyan se dekhne par, n=5 aur usse bade saare integers A mein aayenge (except 6 nahi ban payega, par k=2,l=1 is 7, k=1,l=2 is 8 etc.). Asal mein, linear combination 2k+3l se hum 5,7,8,9,… sab bana sakte hain. 6 nahi ban sakta kyunki k,l≥1.
Step 3: P(B) calculate karna:
P(B)=∑n∈AP(wn)=P(w5)+P(w7)+P(w8)+P(w9)+…
Dhyan dein ki isme n=1,2,3,4,6 missing hain.
P(B)=(n=1∑∞2n1)−(211+221+231+241+261)
P(B)=1−(21+41+81+161+641)=1−(6432+16+8+4+1)=1−6461=643
Correct Option: (1)