Question

A certain player, say$X$ , known to win with probability $0.3$ if the track is fast and $0.4$ if the track is slow. For Monday, there is a $0.7$ probability of a fast track and $0.3$ probability of a slow track. The probability that player $X$ will win a Monday, is
A) $0.22$
B) $0.11$
C) $0.33$
D) None of these

Answer 1

Hint : The question is based on the probability. Probability deals with the occurrence of a random event. For example, when a coin is tossed in the air, the possible outcomes are Head or Tail. In this question the probability of winning in two different tracks and probability of tracks on Monday is given. We need to find the probability i.e., total chance of win in both the tracks on Monday.

Complete step-by-step answer :
For Monday, $P(F) = 0.7$Here we need to determine the probability that player $X$ will win a Monday,
Let us consider,
Player $X$ win on Monday : $W$
Track is Fast : $F$
Track is Slow : $S$
Probability of player $X$ win on Monday : $P(W)$
Probability of player $X$ win in fast track : $P(F)$
Probability of player $X$ win in slow track : $P(S)$
Given,
For Monday, $P(S) = 0.3$
Player $X$ probability to get fast track $P\left( {\dfrac{W}{F}} \right) = 0.3$
Player $X$ probability to get slow track $P\left( {\dfrac{W}{S}} \right) = 0.4$
$P(W) =$ P (to win in fast track OR to win in slow track)
$=$P (to get fast track & to win OR to get slow track & to win)
$= P(W \cap F) + P(W \cap S)$
$= P(F) \times P\left( {\dfrac{W}{F}} \right) + P(S) \times P\left( {\dfrac{W}{S}} \right)$
$= 0.7 \times 0.3 + 0.3 \times 0.4$
$= 0.21 + 0.12$
$P(W) = 0.33$
Hence option C is correct.
Therefore, A certain player, say$X$ , is known to win with probability $0.3$ if the track is fast and $0.4$ if the track is slow. For Monday, there is a $0.7$ probability of a fast track and $0.3$ probability of a slow track. The probability that player $X$ will win a Monday, is $0.33$.
So, the correct answer is “Option C”.

Note : The given question is based on the Multiplication theorem of dependent Events. If two events, A and B are dependent, the probability of occurring 2nd event will be affected by the outcome of the first i.e., $P(A \cap B) = P(A).P(B/A)$.