Probability?

Question

Lawrence, Peter and Richard play a game in which four coins are tossed. Lawrence wins if there is 0 or 1 head. Peter wins if there are 2 heads. Richard wins if there are 3 or 4 heads.

Find the probability that

(ai) Richard will win the game,
Ans: 5/16

(aii) Peter will win the game,
Ans: 3/8

(b) The game is played twice. Find the probability that Lawrence will win one of the two games.
Ans: 55/128

Please show your workings clearly...I'm confused of how to do these questions...Please help..Thanks.

Answer 1

a) Richard win if there are 3 or 4 heads.

the probability of getting a head is 1/2
the probability of getting a tail is 1/2

4 heads.
HHHH or H x H x H x H

(1/2) (1/2) (1/2) (1/2) = 1/16

3 head:
HHHT or H x H x H x T

(1/2) (1/2) (1/2) (1/2)

but the tail does not neccessarily be shown on the first coin, it can be shown on the other coins.
HHHT
HHTH
HTHH
THHH

so there are 4 ways
4 (1/2) (1/2) (1/2) (1/2)
4 (1/16) = 1/4

P (3 or 4 heads) = 1/16 + 1/4
P (3 or 4 heads) = 5/16



Peter wins the game if he gets 2 heads

HHTT or H x H x T x T

(1/2) (1/2) (1/2) (1/2)

there are 4!/(2! 2!) = 6 ways you can rearrange the two heads and the two tails

P (2 heads) = 6 (1/2) (1/2) (1/2) (1/2) = 3/8





First, find the probability the she could win the game. She wins if there is 0 head or 1 head

0 heads means all teals
(1/2)^4 = 1/16

one head
HTTT or H x T x T x T

(1/2) (1/2) (1/2) (1/2)

HTTT
THTT
TTHT
TTTH

4 (1/2) (1/2) (1/2) (1/2) = 1/4

P (0 or 1 head) = 1/4 + 1/16
P = (0 or 1 head) = 5/16

She can either win the the first game and lose the second game

Or she can lose the fist game and win the second game

the chance of losing a game is 1 - 5/16 = 11/16

(5/16) (11/16) + (11/16) (5/16) = 55/128

hope this helps

Answer 2

Lets take into consideration that there are only two sides on a coin so it will give us a 50/50 chance of getting a head or a tail.

a) Richard wins if there are 3 or 4 heads.

the probability of getting a head is 1/2

4 heads.
HHHH or H x H x H x H

(1/2)x (1/2)x (1/2)x (1/2) = 1/16


3 head:
HHHT or H x H x H x T

(1/2)x (1/2)x (1/2)x (1/2) = 1/16


but the tail can be found on different coins as shown:

HHHT
HHTH
HTHH
THHH

so there are 4 arrangements that can be made
4 x(1/2) x(1/2)x (1/2) x(1/2)
4 (1/16) = 1/4

and then add up the no head result:

P (3 or 4 heads) = 1/16 + 1/4
P (3 or 4 heads) = 5/16



b.)Peter wins the game if he gets 2 heads

HHTT or H x H x T x T

(1/2) x(1/2)x (1/2)x (1/2)=1/16

4P2 (4 permuation of 2) = 4!/(2! 2!) = 6

therefore, there are 6 ways you can rearrange the two heads and the two tails

P (2 heads) = 6 (1/2)x(1/2)x(1/2)x(1/2) = 3/8

c.)

Find the probability of winning a game first:

0 head means all tails:

(1/2)x(1/2)x(1/2)x(1/2) = 1/16

one head
HTTT or H x T x T x T

(1/2)x(1/2)x(1/2)x(1/2) = 1/16

HTTT
THTT
TTHT
TTTH

4 ((1/2)x(1/2)x(1/2)x(1/2) = 1/4

P (0 or 1 head) = 1/4 + 1/16
P (0 or 1 head) = 5/16

** this is just the same as the above

1st Combination: Win Loss (win the first game and loss it on the second)
2nd Combination: Loss Win (vice versa of the above)


The chance of losing a game is 1 - (5/16) = 11/16

(5/16)*(11/16) + (11/16)*(5/16) = 55/128

P(1 game out of 2) = 55 /128

Answer 3

Lawrence, Peter and Richard play a game in which four coins are tossed. Lawrence wins if there is 0 or 1 head. Peter wins if there are 2 heads. Richard wins if there are 3 or 4 heads.

Find the probability that

(ai) Richard will win the game,

P(Richard Wins)
= (4C0)[(1/2)^0]*[(1/2)^4] + (4C1)[(1/2)]*[(1/2)³]
= 1*1*(1/16) + 4*(1/2)(1/8) = 1/16 + 4/16 = 5/16

(aii) Peter will win the game,
P(Peter Wins)
= (4C2)[(1/2)²]*[(1/2)²] = 6*(1/4)*(1/4) = 6/16 = 3/8

(b) The game is played twice. Find the probability that Lawrence will win one of the two games.

First calculate the probability that Lawrence will win if one game is played.

P(Lawrence Wins)
= (4C3)[(1/2)³]*[(1/2)] + (4C4)[(1/2)^4]*[(1/2)^0]
= 4*(1/8)*(1/2) + 1*(1/16)*1 = 4/16 + 1/16 = 5/16
______________

Now calculate the probability that if two games are played, Lawrence will win one of the games.

I interpret this to mean he will win at least one game.

P(Lawrence Wins) = 1 - P(Lawrence Loses both games)
= 1 - (1 - 5/16)² = 1 - (11/16)² = 1 - 121/256 = 135/216

If you meant win exactly one game then:

P(Lawrence Wins) = (2C1)(1 - 5/16)(5/16)
= 2(11/16)(5/16) = 110/256 = 55/128

Answer 4

If a coin is flipped four times, determine the probability that it will land heads x times:

Example (two coins): Favorable outcomes: 1 -- HH
Possible outcomes: 4 -- HH, HT, TH, TT
Thus, the probability that the coin will land heads both times is 1 out of 4 .

Pwin(Richard)=(TTTT, HTTT, THTT, TTHT, TTTH)/(TTTT,TTTH,TTHT,THTT,HTTT,TTHH, THHT, HHTT, HTTH, HHHT, THHH, HHHH, HTHH, HHTH, THTH, HTHT)
= 5/16

Pwin(Peter)=(HHTT,THHT,TTHH,HTHT, HTTH, THTH)/(TTTT,TTTH,TTHT,THTT,HTTT,TTHH, THHT, HHTT, HTTH, HHHT, THHH, HHHH, HTHH, HHTH, THTH, HTHT)
=3/8

Pwin(Lawrence)=(HHHT, HHTH, HTHH, THHH, HHHH)/(TTTT,TTTH,TTHT,THTT,HTTT,TTHH, THHT, HHTT, HTTH, HHHT, THHH, HHHH, HTHH, HHTH, THTH, HTHT)
=5/16

Pwin(Lawrence)(one of two games)=(5/16)(11/16)=55/128

(5/16) is one win.
(11/16) is one loss.

Answer 5

is the measure of how likely an event is