any ideas in how to solve this problem ?

Question

A coin is tossed four times. Then it is tossed again, the number of times a head was obtained in the first four tosses. What is the probability of getting at least 5 heads altogether?

Answer 1

Yes I have an idea. First we'll break it down into parts and apply some logic, then find the probability using Pascal's triangle.

The problem:

A coin is tossed four times. Then it is tossed again, the number of times a head was obtained in the first four tosses. What is the probability of getting at least 5 heads altogether?

We know that on 4 coin tosses, we could get 4 heads. That just leaves 1 more head to get on the next 4 tosses in order to get at least 5 heads total. A total of 8 tosses.

We also know that on 4 coin tosses, we could get 3 heads. That leaves 2 heads to get on the next 3 tosses in order to get at least 5 heads total. A total of 7 tosses.

If we get 2 heads, 1 head or 0 heads on the first 4 tosses, then there is no way to get at least 5 heads total. Simply because we would only get 2, 1 or no extra flips. Make sense so far?

Building Pascal's Triangle

We know that for 1 flip we will get either heads or tails, equal chances:

1 flip 1 H.....1 T

To build the triangle, n is the number of flips, and n+1 is the number of values in that row. The first and last numbers in the row are always 1. The rest of the spots are calculated by adding pairs of numbers from the previous row. A number in the row is used as the numerator in the probability fraction, and the sum of all the numbers in the row is used as the denominator. The actual triangle is easier to understand (sorry if the text editor blows the geometry of the image):

1 flip 1 1 sum = 2

2 flips 1 2 1 sum = 4

3 flips 1 3 3 1 sum = 8

4 flips 1 4 6 4 1 sum = 16

5 flips 1 5 10 10 5 1 sum = 32

6 flips 1 6 15 20 15 6 1 sum = 64

7 flips 1 7 21 35 35 21 7 1 sum = 128

8 flips 1 8 28 56 70 56 28 8 1 sum = 256

Does the sum sequence look familiar? Each sum is the previous sum times 2. For n flips, the number of possible outcomes (denominator of the probability fraction) is 2ⁿ,

Now for some more explanation. If we look at the first row:

1 1 represents 1 H and 1 T, or 1 head and 0 heads.

Row 3:

1 3 3 1 represents HHH HHT HTT TTT, or 3 heads, 2 heads, 1 head and 0 heads.

Still with me? We are getting close to the finish.

Row 7 represents:

HHHHHHH HHHHHHT HHHHHTT HHHHTTT HHHTTTT HHTTTTT HTTTTTT TTTTTTT

or

1 7 21 35 35 21 7 1

Since the first 3 entries on this row will give us at least 5 heads, we add those values together:

1+7+21 = 29

Then we divide by the sum of the row:

P = 29/128 = 22.65625%

If we flipped 4 heads on the first 4 tosses, we would use row 8 instead of row 7:

HHHHHHHH HHHHHHHT HHHHHHTT HHHHHTTT HHHHTTTT HHHTTTTT HHTTTTTT HTTTTTTT TTTTTTTT

or

1 8 28 56 70 56 28 8 1

Since the first 4 entries on this row will give us at least 5 heads, we add those values together:

1+8+28+56 = 93

Divide by the sum of the row:

P = 93/256 = 36.328125%

You have a higher probability of making 5 heads based on how well you do in the first 4 flips, 3H or 4H.