axe blue cends and 1 ved ard in a box. Albest ans Bemord teke tuns to dyaw a ce

Question

axe blue cends and 1 ved ard in a box. Albest ans Bemord teke tuns to dyaw a ce om ftom the box, with Albest being the first Fo st pesson who dqus the fed card will dYaw. The fixst gome. Define X to be num bey of card dyaws untila winney em esges. a) If the e coxds are daun without veplacement, a- find the PMF of X. f the cards ave dsaw without Yeplacem ent what is the pobebility hat Besnayd wvill uin? E the cards axe da with replacement, Find the PMF of X. d If the ceds are drau with replacement, wha+ is the probability the Benard ill win? ven thot Benerd wins, devive the eicbabi li t perating function for the num ber of dva w i+ took hin to win

Explanation / Answer

(a) Here total number of cards = 7

Red card = 1

Here cards are drawn without replacemnt.

p(x) = 1/7 ; X = 1,2,3,4,5,6,7

[ Note : Here it can be calculated as like if 2 is the number of card draws when red comes.

so, Pr(X = 2) = Pr(first card is not red) * Pr(second caard is red) = 6/7 * 1/6 = 1/7

similarly, for each probability it would be same. ]

so, p(x) = 1/7 ; X = 1,2,3,4,5,6,7

(b) HEre the probability that Benard will win. Bernanrd will win when X would be 2,4 & 6.

Pr( BErnanrd will win) = 3 * 1/7 = 3/7

(c) If the cards are drawn with replacement, PMF of X. The given distribution is geometric distribution.

p = 1/7

p(x) = (6/7)X-1 (1/7)

(d) Here the probability that bernanrd will win. Bernanrd will win when the X would be even number so the given distribution is an infinity distribution.

Pr(BErnanrd will win) = 1/7 [6/7 + (6/7)3 + .....] = 1/7 [(6/7)/ (1 - 36/49)]= 6/49 * (49/13) = 6/13

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