Probability

Question

4 fair coins are simultaneously flipped and 1 fair 4-sided die (with numbers 1-4 on the faces) is tossed. What is the probability that the number of heads is equal to the number that comes up on the die? Show all steps to answer for credit.

Answer 1

 

Hi Chris,

 

4 different possibilities we need to consider are

 

   a) # of Heads = 1 & 1 comes up on the die,

 

   b) # of Heads = 2 & 2 comes up on the die,

 

   c) # of Heads = 3 & 3 comes up on the die, and

 

   d) # of Heads = 4 & 4 comes up on the die.

 

In case of a), the probability is

 

   {₄C₁ · (1/2)¹ · (1/2)³} · (1/4) = 1/16

 

   where _xC_y = x!/((x-y)!·y!) is the combinatorial operation.

 

In case of b), the probability is
 
   {₄C₂ · (1/2)² · (1/2)²} · (1/4) = 3/32.

 

In case of c), the probability is

   {₄C₃ · (1/2)³ · (1/2)¹} · (1/4) = 1/16.

 

In case of d), the probability is

   {₄C₄ · (1/2)⁴ · (1/2)⁰} · (1/4) = 1/64.

 

So, the answer is 15/64 = 0.2344 (= (1/16) + (3/32) + (1/16) + 1/64).

 

 

 

Answer 2

Coins (number of heads)

 

0 = 1/16

1 = 4/16

2 = 6/16

3 = 4/16

4 = 1/16

 

Die

 

1 = 1/4

2 = 1/4

3 = 1/4

4 = 1/4

 

Both 1:  4/16 * 1/4 = 4/64

Both 2:  6/16 * 1/4 = 6/64

Both 3:  4/16 * 1/4 = 4/64

Both 4:  1/16 * 1/4 = 1/64

 

Total:  15/64

 

 

 

Answer 3

Each flip of a coin is an independent event, and the fact that four coins were flipped is of no significance. The odds are 1 in 2 for each coin and each flip.

 

So, the chance of getting four heads will be ½ x ½ x ½ x ½ = 1/16 = 0.0625, or 6.25%

 

Even though the die has numbers on only four sides, there are six sides, so the odds of rolling a four are 1 in 6 = 0.1667 = 16.67%

 

Combining the two events, we get 0.0625 x 0.1667 = 0.0104 = 1.04%

Answer 4

Chris,

 

You have - for each coin - 2 possible ways

 

 

1 - HHHH

2 - HHHT   ANOTHER EIGHT FOR T AND YOU HAVE 16 WAYS JUST FOR THE COIN. AFTER YOU KNOW WHAT YOU ARE

3 - HHTH   DOING, YOU WOULDN'T NEED TO LIST THEM LIKE THIS. YOU CAN JUST SAY 2 * 2 * 2 * 2 = 16. WITH THE  

4 - HHTT   2S REPRESENTING THE TWO POSSSIBLE OUTCOME FOR EACH (H, T)

5 - HTHH

6 - HTHT

7 - HTTH

8 - HTTT

T.....

THE IDE HAS 4 OUTCOMES WHICH GOES WITH ANY OF THE 16 POSSIBLE OUTCOMES OF THE 4 COINS. SO, YOU HAVE 16*4 = 64 POSSIBLE COUTCOMES.

NOW, IF THE DIE TURNS UP 4 HEADS - THE EASIEST CASE - YOU WILL HAVE HHHH FOR THE COINS. 1 - CASE

IF THE DIE TURNS UP 1 HEADS - YOU WILL HAVE HTTT, THTT, TTHT, TTTH FOR THE COINS. 4 - CASES

YOU DO THE SAME FOR 2 AND 3, ADD UP ALL THE FOUR CASES AND DIVIDE THEM BY 64 (ALL POSSIBLE OUTCOMES). IF YOU CAN'T DO THE TALLING UP FOR 2 AND 3, TEXT ME AND I WILL HELP WITH IT.