Four cards will be dealt off the top of a well-shuffled deck. There are two options. (i) To win $10 if the first card is club and the second is a diamond and the third is a heart and the fourth is a spade. (ii) To win $10 if the four cards are of four different suits. Compute the probability of winning $10 for each case. Which option is better

0.0043957 ; 0.105498 ; 2nd option is better

Step-by-step explanation:

Option 1:

First card :P(club) = 13/52

Second card: P(diamond) = 13/51

Third card : P(heart) = 13 / 50

Fourth card: P(spade) = 13/49

Hence,

P(club, diamond, heart, spade) = (13/52) * (13/51) * (13/50) * (13/49) = 0.0043957

Probability of winning $10 = 0.0043957

2nd option:

P(card of different suit) [order of arrangement does not matter]

(13C1 * 13C1 * 13C1 * 13C1) / 52C4

= 28561 / 270725

= 0.105498

Probability of winning $10 = 0.105498

Because the probability of winning for the second option is higher than for the first option, then the second option is better.

b

step-by-step explanation:

a precentage of 6th graders. the population is all 6th graders. let me know if i'm wrong, i hope this ! : )

step-by-step explanation: