Question

Suppose you toss a fair coin until you’ve gotten at heads at least twice or tails at least four times (not necessarily consecutive), and then you stop. What is the probability that your last coin toss came up tails?

Answer 1

Answer: The probability of getting a tail at the last toss is 27%

Explanation:

The probability of getting a tail in your last coin toss is dependent on how many possible ways the out come of the tail can appear in all the toss, which tail being the last outcome is one of the possible ways.

CALCULATE THE POSSIBLE WAY A TAIL CAN APPEAR IN THE TOSS:

Since it's assume we have 2 head and 4 tails that means the total number of toss is

2+4 =6 =n

using combination formula

nCr = n! ÷ (r! (n - r)!)

The arrangements of tail becomes

6C4

therefore;

6! ÷ ( 4! × (6 - 4)!)

6! ÷ (4! × 2!)

6! = 6×5×4×3×2×1 = 720

4! = 4×3×2×1 = 24

2! = 2×1 = 2

Therefore;

720 ÷ (24 × 2) = 720÷48 = 15

Therefore they are 15 possible ways 4 tails can appear in the 6 toss of the coin, if the head does not appear consecutively.

WHAT IS THE PROBABILITY OF GETTING A TAIL AT THE LAST TOSS:

Since they are 15 possible ways 4 tail can appear without the head appearing consecutively. That means the probability of having a tail at the last toss will be;

4/15 = 0.27

Therefore the tail has 27% probability of occuring at the last toss if the head does not occur consecutively.

Answer 2

Explanation:

Given,

To give the probability that your last coin toss came up tails

the probability that the last coin toss came up tails = likelihood of occurring that all 4 were tails when coin tossed 4 times + likelihood of occurring last toss gives tail & 1st four tosses were head when coin is tossed 5 times

= (1/2)^4 + [(1/2)^5 + (1/2)^5 + (1/2)^5 + (1/2)^5]

= 1/16 + [4*(1/32)]

= 1/16 + 1/8

= 3/16

= 0.1875