Question

A box contains 2 identical pistols. Pistol A contains 7 live bullets and 3 blank bullets while pistol B contains 3 live bullets and 7 blanks. Following a particularly annoying question in his Stat 230 class, evil Professor Moriarty chooses a pistol at random and then fires once directly at Holmes.

a. Find the probability that Holmes survives this shot (i.e. a blank bullet is fired).

b. If Holmes survives, he takes the other pistol and fires directly at Moriarty. What is the probability that Holmes survives the first shot and Moriarty survives this second shot? (i.e. a both bullets fired are blanks.).

c. If this ill-considered game continues until one of the two is shot with a live bullet, what is the probability that the person who survives is Holmes?

Answer 1

Here there are 2 identical pistols.

Pistol A contains 7 live bullets and 3 blank bullets. Pistol B contains 3 live bullets and 7 blanks.

(a) P(Selecting any one of the pistol) = 1/2

P(Holmes survives the shot) = P(Moriarity Choose pistol 1) * P(Fired a blank shot) + P(Moriarity chooses pistol 2) * P(Fired a blank shot)

= 1/2 * 0.7 + 1/2 * 0.3 = 0.5

(b) Here there are two conditions

(i) Moriarty fired first pistol and fired blank and Holmes fired second pistol and fired blank.

Probability of that event is = 3/10 * 7/10 = 21/100 = 0.21

(ii)  Moriarty fired second  pistol and fired blank and Holmes fired first pistol and fired blank.

Probability of that event is = 7/10 * 3/10 = 21/100 = 0.21

P(Moriarty selects any random pistol) = 1/2

so,

P(BOth survivies in that condition) = 1/2 * 0.21 + 1/2 * 0.21 = 0.21

(c) Here we have to find the probability that moriarity dies and holmes survives.

There are two condition:

(i) Moriarity chooses first pistol

(1) Moriarity get killed in first attempt = P(Moriarity fired blank) * P(Holmes Fired bullet) = 3/10 * 3/10 = 9/100 = 0.09

(2) Moriarity get killed in second attempt = P(M Fired blank in first attempt) * P(H Fired blank also) * P(M Fired blank again) * P(Holmes Fired bullet in second attempt) = 3/10 * 7/10 * 2/9 * 3/9 = 0.0156

(3) Moriarity get killed in third attempt = 3/10 * 7/10 * 2/9 * 6/9 * 1/8 * 3/8 = 0.001458

Innext attempt moriarity will defenitely kill holmes as there is no chance of firing blank now.

P(Holmes survivies if moriairty chooses first pistol) = 0.09 + 0.0156 + 0.001458 = 0.1070

(ii) Moririty chooses second pistol, here we will find the probability that holmes get killed as after three rounds moriarty definitely get killed by holmes

(1) Holmes get killed in first attempt = P(Moriarity Fired bullet) ) = 3/10

(2) Holmes  get killed in second attempt = P(M Fired Blank) * P(H Fired blank) * P(M Fired Bullet) = 7/10 * 3/10 * 3/9 = 0.07

(3) Holmes get killed in third attempt = 7/10 * 3/10 * 6/9 * 2/9 * 3/8 = 0.01167

(4) Holmes get killed in fourth attempt = 7/10 * 3/10 * 6/9 * 2/9 * 3/8 * 1/8 * 3/7 = 0.000625

In next attempt Holmes will defenitely kill moriarity as there is no chance of firing blank now.

P(Holmes get killed if chosend second pistol) = 3/10 + 0.07 + 0.01167 + 0.000625 = 0.3823

P(Moriraty get killed if choosedn second pistol) =1 - 0.3823 = 0.6177

P(Moriraty get killed and holmes survives) = 1/2 * 0.1070 + 1/2 * 0..6177 = 0.3624