A coin is tossed 8 times. If the probability that exactly 4 heads appear in the first six tosses and exactly 3 heads appear in the last five tosses is p, then 96p is equal to ____ .
Answer: 9
Solutions · 1
Official solution
Let the outcomes of the 8 tosses be denoted by X1,X2,X3,X4,X5,X6,X7,X8.
The condition ``exactly 4 heads in the first six tosses'' means the block (x1,x2,x3,x4,x5,x6) contains 4 heads.
The condition ``exactly 3 heads in the last five tosses'' means the block (x4,x5,x6,x7,x8) contains 3 heads.
Notice that these two blocks overlap in the common tosses (x4,x5,x6).
Let k be the number of heads in the common tosses (x4,x5,x6).
Then:
In (x1,x2,x3) we must have 4−k heads (so that the first six have total 4 heads).
In (x7,x8) we must have 3−k heads (so that the last five have total 3 heads).
Now we count the number of valid outcome sequences for each possible value of k.
k=1 \therefore ways to overlap = 3C1
Here, we choose:
3C3 ways to put 4−k=3 heads in (x1,x2,x3),
3C1 ways to put k=1 head in (x4,x5,x6),
2C2 ways to put 3−k=2 heads in (x7,x8).
3C3×3C1×2C2=3k=2
Here, we choose:
3C2 ways to put 4−k=2 heads in (x1,x2,x3),
3C2 ways to put k=2 heads in (x4,x5,x6),
2C1 ways to put 3−k=1 head in (x7,x8).
3C2×3C2×2C1=18k=3
Here, we choose:
3C1 ways to put 4−k=1 head in (x1,x2,x3),
1 way to put k=3 heads in (x4,x5,x6) (all heads),
1 way to put 3−k=0 heads in (x7,x8) (all tails).
3C1×1=3
Total favourable outcomes =3+18+3=24.
Since each full sequence of 8 tosses has probability (21)8, we get:
Total ways 24(21)8
So p=24(21)8=2824=25624=323, hence 96p=96⋅323=9.
96×2824=166×24=9