Question

5 cards are drawn successively from a well-shuffled pack of 52 cards with replacement. Determine the probability that (i) all the five cards should be spades? (ii) only 3 cards should be spades? (iii) none of the cards is a spade?​

Answer 1

Answer:

The light sensitivity of the silver halides is key to the photographic process. Tiny crystals of all three of these compounds are used in making photographic film. ... When film containing Ag+ and Cl- is exposed to light energy, the chlorine ion's extra electron is ejected and then captured by a silver ion.

Answer 2

$\tt Let\:X\:be\:the\:number\:of\:spade\:cards \\ among\:5\:cards \\ \\ \tt Since\: drawing\:of\:cards\:is\:with\:replacement\:the\:trails \\ \tt are\:Bernoullis\:trails \\ \\ \tt In\:a\:well\:shuffled\:pack\:there\:are \\ \tt 13\:spade\:cards \\ \\ \tt \implies p = \frac{13}{52} = \frac{1}{4} \\ \\ \tt \implies q = 1 - p = 1 - \frac{1}{4} = \frac{3}{4} \\ \\ \tt We\:know\:that \\ \\ \tt {P}_{(X=x)} = {}^{n}{C}_{x}.{p}^{x}.{q}^{n-x} \\ \\ \tt (i) {P}_{(all\:cards\:are\:spades)} \\ \\ \tt \implies {P}_{(X=5)} = {}^{5}{C}_{5}.{(\frac{1}{4})}^{5}.{(\frac{3}{4})}^{0} \\ \\ \tt \implies {P}_{(X=5)} = \frac{1}{{4}^{5}} \\ \\ \tt \implies {P}_{(X=5)} = \frac{1}{1024} \\ \\ \tt \therefore \boxed{\bold{\underline{\red{\tt Probability\:of\:all\:cards\:of\:spades\:is\: \frac{1}{1024} }}}} \\ \\ \tt (ii) {P}_{(3\:cards\:should\:be\:spades)} \\ \\ \tt \implies {P}_{(X=3)} = {}^{5}{C}_{3}.{(\frac{1}{4})}^{3}.{(\frac{3}{4})}^{2} \\ \\ \tt \implies {P}_{(X=3)} = \frac{5!}{2!.3!} \frac{1}{{4}^{3}}.\frac{{3}^{2}}{{4}^{2}} \\ \\ \tt \implies {P}_{(X=3)} = \frac{9×10}{{4}^{5}} \\ \\ \tt \implies {P}_{(X=3)} = \frac{45}{512} \\ \\ \tt \therefore \boxed{\bold{\underline{\red{\tt Probability\:of\: only\:3\:cards\:are\:of\:spade\:is\: \frac{45}{512} }}}} \\ \\ \tt (iii) {P}_{(none\:card\:is\:of\:spade)} \\ \\ \tt \implies {P}_{(X=0)} = {}^{5}{C}_{0}.{(\frac{1}{4})}^{0}.{(\frac{3}{4})}^{5} \\ \\ \tt \implies {P}_{(X=0)} = \frac{{3}^{5}}{{4}^{5}} \\ \\ \tt \implies {P}_{(X=0)} = \frac{243}{1024} \\ \\ \tt \therefore \boxed{\bold{\underline{\red{\tt Probability\:of\:none\:card\:is\:of\:spade\:is\: \frac{243}{1024} }}}}$