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Coin Tosses Overlapping Head Counts: 96p = 9 | JEE Main 2026

JEE Maths question with a full step-by-step solution.

Question
A coin is tossed 88 times. If the probability that exactly 44 heads appear in the first six tosses and exactly 33 heads appear in the last five tosses is pp, then 96p96p is equal to
Solution
Answer: 9 (± 0.01)
Step 1: Label the tosses x1,…,x8x_1,\ldots,x_8. "First six" is x1…x6x_1\ldots x_6; "last five" is x4…x8x_4\ldots x_8; the common tosses are x4,x5,x6x_4,x_5,x_6. Let kk = number of heads among the common three. Then x1x2x3x_1x_2x_3 has 4−k4-k heads and x7x8x_7x_8 has 3−k3-k heads. Step 2: Count for each kk (number of head-arrangements):
k=1: 3C3⋅3C1⋅2C2=1⋅3⋅1=3,k=1:\ {}^3C_3\cdot{}^3C_1\cdot{}^2C_2=1\cdot3\cdot1=3,
k=2: 3C2⋅3C2⋅2C1=3⋅3⋅2=18,k=2:\ {}^3C_2\cdot{}^3C_2\cdot{}^2C_1=3\cdot3\cdot2=18,
k=3: 3C1⋅3C3⋅2C0=3⋅1⋅1=3.k=3:\ {}^3C_1\cdot{}^3C_3\cdot{}^2C_0=3\cdot1\cdot1=3.
 Total =3+18+3=24=3+18+3=24. Step 3: Each specific outcome has probability (12)8\left(\tfrac12\right)^8, so p=24256p=\dfrac{24}{256}. Then
96p=96⋅24256=6⋅2416=9.96p=96\cdot\frac{24}{256}=\frac{6\cdot24}{16}=9.
 Correct answer: 9
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